diff options
Diffstat (limited to '647')
193 files changed, 5176 insertions, 0 deletions
diff --git a/647/CH1/EX1.1/Example1_1.sce b/647/CH1/EX1.1/Example1_1.sce new file mode 100755 index 000000000..7ec57d2f1 --- /dev/null +++ b/647/CH1/EX1.1/Example1_1.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 1.1
+// Page: 4
+
+// Solution
+
+printf("Example: 1.1 - Page: 4\n\n");
+
+//*****Data*****//
+g_Earth = 9.83;// [m/square s]
+F_Earth = 800;// [N]
+g_Moon = 3.2;// [m/square s]
+//************//
+
+// From the expression of force, the force on the man on the Eath's surface is given by:
+// F = m*g_Earth
+m = F_Earth/g_Earth;// [kg]
+
+// On the moon, the weight of the mass is equal to the force acting on the mass on the moon and is given by
+F_Moon = m*g_Moon;// [N]
+
+printf("Weight of the man on the moon is %f N\n",F_Moon);
\ No newline at end of file diff --git a/647/CH1/EX1.10/Example1_10.sce b/647/CH1/EX1.10/Example1_10.sce new file mode 100755 index 000000000..3519adc6f --- /dev/null +++ b/647/CH1/EX1.10/Example1_10.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 1.10
+// Page: 12
+
+// Solution
+
+printf("Example: 1.10 - Page: 12\n\n");
+
+//*****Data*****//
+m = 1200;// [kg]
+v1 = 10;// [km/h]
+v2 = 100;// [km/h]
+time = 1;// [min]
+//***************//
+
+v1 = 10*1000/3600;// [m/s]
+v2 = 100*1000/3600;// [m/s]
+W = (1/2)*m*(v2^2 - v1^2);// [J]
+time = time*60;// [s]
+P = W/time;// [W]
+printf("Power required is %.2f kW\n",P/1000);
\ No newline at end of file diff --git a/647/CH1/EX1.11/Example1_11.sce b/647/CH1/EX1.11/Example1_11.sce new file mode 100755 index 000000000..e1bef5e98 --- /dev/null +++ b/647/CH1/EX1.11/Example1_11.sce @@ -0,0 +1,40 @@ +clear;
+clc;
+
+// Example: 1.11
+// Page: 13
+
+printf("Example: 1.11 - Page: 13\n\n");
+
+//*****Data*****//
+dia = 0.3;// [m]
+m = 100;// [kg]
+P_atm = 1.013*10^5;// [N/square m]
+g = 9.792;// [m/square s]
+//**************//
+
+Area = (%pi/4)*dia^2;// [square m]
+//Solution (a)(i)
+// Force exerted by the atmosphere:
+F_atm = P_atm*Area;// [N]
+// Force exerted by piston & metal block:
+F_mass = m*g;// [N]
+// Total force acting upon the gas:
+F = F_atm + F_mass;// [N]
+printf("Total Force eacting upon the gas is %.1f N\n",F);
+
+// Solution (a)(ii)
+Pressure = F/Area;// [N/square m]
+printf("Pressure exerted is %.3f kPa\n\n",Pressure/1000);
+
+// Solution (b)
+// The gas expands on application of heat, the volume of the gas goes on increasing and the piston moves upward.
+Z = 0.5;// [m]
+// Work done due to expansion of gas:
+W = F*Z;// [J]
+printf("Work due to expansion by the gas is %.3f kJ\n\n",W/1000);
+
+// Solution (c)
+// Change in potential energy of piston and weight after expansion process:
+Ep = m*g*Z;// [J]
+printf("Change in Potential Energy is %.1f J\n",Ep);
\ No newline at end of file diff --git a/647/CH1/EX1.12/Example1_12.sce b/647/CH1/EX1.12/Example1_12.sce new file mode 100755 index 000000000..e9dcbc162 --- /dev/null +++ b/647/CH1/EX1.12/Example1_12.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+
+// Example: 1.12
+// Page: 24
+
+printf("Example: 1.12 - Page: 24\n\n");
+
+// Solution
+
+// The relation is:
+// (C/5) = ((F - 32)/9)
+// For C = F
+C = - (32*5/4); // [degree Celsius]
+printf("The temperature which has the same value on both the centigrade and Fahrenheit scales is %d degree Celsius or %d degree Fahrenheit\n",C,C);
\ No newline at end of file diff --git a/647/CH1/EX1.13/Example1_13.sce b/647/CH1/EX1.13/Example1_13.sce new file mode 100755 index 000000000..15431a639 --- /dev/null +++ b/647/CH1/EX1.13/Example1_13.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 1.13
+// Page: 24
+
+printf("Example: 1.13 - Page: 24\n\n");
+
+// Solution
+
+//*****Data*****//
+delta_T_C = 30;// [OC]
+//*************//
+
+// The relation between the Kelvin temperature scale and the Celsius temperature scale:
+// T(K) = T(OC) + 273.15
+// Here, the temperature rise is to be expressed in terms of K, but the difference in temperature will be the same in the Kelvin and Celsius scales of temperature:
+delta_T_K = delta_T_C;// [K]
+printf("The rise in temperature in the Kelvin scale is %d K\n",delta_T_K);
+// The emperical relationship between the Rankine and Kelvin scales is given by:
+delta_T_R = 1.8*delta_T_K;// [R]
+printf("The rise in temperature in the Rankine scale is %d R\n",delta_T_R);
+delta_T_F = delta_T_R;// [OF]
+printf("The rise in temperature in the Fahrenheit scale is %d OF\n",delta_T_F);
\ No newline at end of file diff --git a/647/CH1/EX1.2/Example1_2.sce b/647/CH1/EX1.2/Example1_2.sce new file mode 100755 index 000000000..90116befd --- /dev/null +++ b/647/CH1/EX1.2/Example1_2.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 1.2
+// Page: 5
+
+// Solution
+
+printf("Example: 1.2 - Page: 5\n\n");
+
+//*****Data*****//
+m1 = 1.5;// [mass of the body, kg]
+m2 = 6*10^(24);// [mass of the Earth, kg]
+G = 6.672*10^(-11);// [N.square m/square.kg]
+r = 6000*10^(3);// [m]
+//************//
+
+// According to Newton's universal law of gravity:
+F = G*m1*m2/r^2;// [N]
+printf("Gravitational force on the body is %.2f N\n",F);
\ No newline at end of file diff --git a/647/CH1/EX1.3/Example1_3.sce b/647/CH1/EX1.3/Example1_3.sce new file mode 100755 index 000000000..e8d75be73 --- /dev/null +++ b/647/CH1/EX1.3/Example1_3.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 1.3
+// Page: 5
+
+// Solution
+
+printf("Example: 1.3 - Page: 5\n\n");
+
+//*****Data*****//
+r_Moon = 0.3;// [km]
+r_Earth = 1;// [km]
+m2 = 1;// [mass of body, kg]
+mMoon_By_mEarth = 0.013;// [kg/kg]
+//***************//
+
+// According to the Newton's universal law of gravitation:
+Fe_By_Fm = (1/mMoon_By_mEarth)*(r_Moon/r_Earth)^2;
+printf("Mass of 1 kg will weigh %.2f kg on moon\n",Fe_By_Fm);
\ No newline at end of file diff --git a/647/CH1/EX1.4/Example1_4.sce b/647/CH1/EX1.4/Example1_4.sce new file mode 100755 index 000000000..78eacce40 --- /dev/null +++ b/647/CH1/EX1.4/Example1_4.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 1.4
+// Page: 6
+
+// Solution
+
+printf("Example: 1.4 - Page: 6\n\n");
+
+//*****Data*****//
+h = 40;// [cm]
+density = 14.02;// [g/cubic cm]
+g = 9.792;// [m/square s]
+//*************//
+
+P = h*density*g/1000;// [N/square cm]
+P = P*10;// [kPa]
+
+printf("The absolute pressure is %.3f kPa\n",P);
\ No newline at end of file diff --git a/647/CH1/EX1.5/Example1_5.sce b/647/CH1/EX1.5/Example1_5.sce new file mode 100755 index 000000000..9061c5c6f --- /dev/null +++ b/647/CH1/EX1.5/Example1_5.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 1.5
+// Page: 7
+
+// Solution
+
+printf("Example: 1.5 - Page: 7\n\n");
+
+//*****Data*****//
+Patm = 112;// [kPa]
+density = 1200;// [kg/cubic m]
+g = 9.81;// [m/sqaure s]
+h = 0.62;// [m]
+//**************//
+
+P = Patm + (density*g*h/1000);// [kPa]
+
+printf("The absolute pressure within the container is %.3f kPa\n",P);
\ No newline at end of file diff --git a/647/CH1/EX1.6/Example1_6.sce b/647/CH1/EX1.6/Example1_6.sce new file mode 100755 index 000000000..805672ba3 --- /dev/null +++ b/647/CH1/EX1.6/Example1_6.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Example: 1.6
+// Page: 9
+
+// Solution
+
+printf("Example: 1.6 - Page: 9\n\n");
+
+//*****Data*****//
+F = 150;// [N]
+Displacement = 10;// [m]
+//**************//
+
+W = F*Displacement;// [J]
+
+printf("Work done by the system is %d J\n",W);
\ No newline at end of file diff --git a/647/CH1/EX1.7/Example1_7.sce b/647/CH1/EX1.7/Example1_7.sce new file mode 100755 index 000000000..402067119 --- /dev/null +++ b/647/CH1/EX1.7/Example1_7.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 1.7
+// Page: 9
+
+// Solution
+
+printf("Example: 1.7 - Page: 9\n\n");
+
+//*****Data*****//
+P = 560*10^3;// [Pa]
+Vinit = 3;// [cubic m]
+Vfinal = 5;// [cubic m]
+Wext = 210*10^3;// [J]
+//*************//
+
+W = P*(Vfinal - Vinit);// [J]
+// Again the system receives 210 kJ of work from the external agent.
+W = W - Wext;// [J]
+
+printf("Actual Work done by the system is %.1e J\n",W);
\ No newline at end of file diff --git a/647/CH1/EX1.8/Example1_8.sce b/647/CH1/EX1.8/Example1_8.sce new file mode 100755 index 000000000..a4906635a --- /dev/null +++ b/647/CH1/EX1.8/Example1_8.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 1.8
+// Page: 11
+
+// Solution
+
+printf("Example: 1.8 - Page: 11\n\n");
+
+//*****Data*****//
+g = 9.81;// [m/square s]
+Z = 100;//[m]
+//***************//
+
+// Basis: 1 kg of water
+m = 1;// [kg]
+Ep = m*g*Z;// [J]
+printf("Change in potential Energy is %d J\n",Ep)
\ No newline at end of file diff --git a/647/CH1/EX1.9/Example1_9.sce b/647/CH1/EX1.9/Example1_9.sce new file mode 100755 index 000000000..4fe29b682 --- /dev/null +++ b/647/CH1/EX1.9/Example1_9.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 1.9
+// Page: 11
+
+// Solution
+
+printf("Example: 1.9 - Page: 11\n\n");
+
+//*****Data*****//
+m = 15; // [kg]
+g = 9.81;// [m/square s]
+V1 = 0;// [m/square s]
+Z1 = 12;// [m]
+Z2 = 0;// [m]
+//***************//
+
+// At initial condition, V1 = 0, so kinetic energy is zero.
+// At final condition, Z2 = 0, so potential energy is zero.
+// Ep1 + Ek1 = Ep2 + Ek2
+deff('[y] = f(V2)','y = ((1/2)*m*V1^2) + (m*g*Z1) - (((1/2)*m*V2^2) + (m*g*Z2))');
+V2 = fsolve(7,f);
+
+printf("The velocity of the metal block is %.2f m/s\n",V2);
+
+Ek2 = (1/2)*m*V2^2;// [J]
+printf("The final Kinetic Energy is %.1f J\n",Ek2);
\ No newline at end of file diff --git a/647/CH10/EX10.1/Example10_1.sce b/647/CH10/EX10.1/Example10_1.sce new file mode 100755 index 000000000..3c7437b8b --- /dev/null +++ b/647/CH10/EX10.1/Example10_1.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 10.1
+// Page: 390
+
+printf("Example: 10.1 - Page: 390\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 10.1 on page number 390 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 10.1 on page 390 of the book.");
\ No newline at end of file diff --git a/647/CH10/EX10.10/Example10_10.sce b/647/CH10/EX10.10/Example10_10.sce new file mode 100755 index 000000000..1a0644d94 --- /dev/null +++ b/647/CH10/EX10.10/Example10_10.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 10.10
+// Page: 412
+
+printf("Example: 10.10 - Page: 412\n\n");
+
+// Solution
+
+//*****Data******//
+P = 101.3;// [kPa]
+P1sat = 100.59;// [kPa]
+P2sat = 99.27;// [kPa]
+x1 = 0.532;
+//****************//
+
+x2 = 1 - x1;
+gama1 = P/P1sat;
+gama2 = P/P2sat;
+A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))^2;
+B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))^2;
+
+// For solution containing 10 mol percent benzene:
+x1 = 0.10;
+x2 = 1 - x1;
+gama1 = exp(A/(1 + (A*x1/(B*x2))^2));
+gama2 = exp(B/(1 + (B*x2/(A*x1))^2));
+printf("Activity Coeffecient\n");
+printf("gama1 = %.3f\n",gama1);
+printf("gama2 = %.3f\n",gama2);
\ No newline at end of file diff --git a/647/CH10/EX10.11/Example10_11.sce b/647/CH10/EX10.11/Example10_11.sce new file mode 100755 index 000000000..ccb833502 --- /dev/null +++ b/647/CH10/EX10.11/Example10_11.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Example: 10.11
+// Page: 413
+
+printf("Example: 10.11 - Page: 413\n\n");
+
+// Solution
+
+//*****Data******//
+P = 760;// [mm of Hg]
+P1sat = 995;// [mm of Hg]
+P2sat = 885;// [mm of Hg]
+x1 = 0.335;
+T = 64.6;// [OC]
+//****************//
+
+x2 = 1 - x1;
+gama1 = P/P1sat;
+gama2 = P/P2sat;
+A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))^2;
+B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))^2;
+
+// For solution containing 11.1 mol percent acetone:
+x1 = 0.111;
+x2 = 1 - x1;
+gama1 = exp((A*x2^2)/(x2 + (A*x1/(B))^2));
+gama2 = exp((B*x1^2)/(x1 + (B*x2/(A))^2));
+y1 = 1/(1 + (gama2*x2*P2sat/(gama1*x1*P1sat)));
+y2 = 1 - y1;
+printf("Equilibrium Composition\n");
+printf("Acetone Composition = %.2f %%\n",y1*100);
+printf("Chloform composition = %.2f %%\n",y2*100);
\ No newline at end of file diff --git a/647/CH10/EX10.12/Example10_12.sce b/647/CH10/EX10.12/Example10_12.sce new file mode 100755 index 000000000..522e83e4e --- /dev/null +++ b/647/CH10/EX10.12/Example10_12.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 10.12
+// Page: 414
+
+printf("Example: 10.12 - Page: 414\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 10.12 on page number 414 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 10.12 on page 414 of the book.");
\ No newline at end of file diff --git a/647/CH10/EX10.13/Example10_13.sce b/647/CH10/EX10.13/Example10_13.sce new file mode 100755 index 000000000..90b1811dc --- /dev/null +++ b/647/CH10/EX10.13/Example10_13.sce @@ -0,0 +1,54 @@ +clear;
+clc;
+
+// Example: 10.13
+// Page: 418
+
+printf("Example: 10.13 - Page: 418\n\n");
+
+// Solution
+
+//*****Data******//
+// 1: iso - butanol
+// 2: iso - propanol
+T = 50 + 273;// [K]
+x1 = 0.3;
+V1 = 65.2;// [cubic cm/mol]
+V2 = 15.34;// [cubic cm/mol]
+// For Wilson equation:
+a12 = 300.55;// [cal/mol]
+a21 = 1520.32;// [cal/mol]
+// For NTRL equation:
+b12 = 685.21;// [cal/mol]
+b21 = 1210.21;// [cal/mol]
+alpha = 0.552;
+R = 2;// [cal/mol K]
+//******************//
+
+x2 = 1 - x1;
+// A: Estimation of activity coeffecient using Wilson equation:
+// From Eqn. 10.65:
+A12 = (V2/V1)*exp(-a12/(R*T));
+// From Eqn. 10.66:
+A21 = (V1/V2)*exp(-a21/(R*T));
+// From Eqn. 10.67:
+gama1 = exp(-log(x1 + A12*x2) + x2*((A12/(x1 + A12*x2)) - (A21/(A21*x1 + x2))));
+// From Eqn. 10.68:
+gama2 = exp(-log(x2 + A21*x1) - x1*((A12/(x1 + A12*x2)) - (A21/(A21*x1 + x2))));
+printf("Wilson equation\n");
+printf("Activity Coeffecient of iso - butanol is %.3f\n",gama1);
+printf("Activity Coeffecient of iso - propanol is %.3f\n",gama2);
+printf("\n");
+
+// A: Estimation of activity coeffecient using NTRL equation:
+t12 = b12/(R*T);
+t21 = b21/(R*T);
+G12 = exp(-alpha*t12);
+G21 = exp(-alpha*t21);
+// From Eqn. 10.70:
+gama1 = exp((x2^2)*(t21*(G21/(x1 + x2*G21))^2 + (t12*G12/(x2 + x1*G12)^2)));
+// From Eqn. 10.71:
+gama2 = exp((x1^2)*(t12*(G12/(x2 + x1*G12))^2 + (t21*G21/(x1 + x2*G21)^2)));
+printf("NTRL equation\n");
+printf("Activity Coeffecient of iso - butanol is %.3f\n",gama1);
+printf("Activity Coeffecient of iso - propanol is %.3f\n",gama2);
\ No newline at end of file diff --git a/647/CH10/EX10.14/Example10_14.sce b/647/CH10/EX10.14/Example10_14.sce new file mode 100755 index 000000000..dc9b31eb6 --- /dev/null +++ b/647/CH10/EX10.14/Example10_14.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+// Example: 10.14
+// Page: 426
+
+printf("Example: 10.14 - Page: 426\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.4;// [mole fraction of ethane in vapour phase]
+x2 = 0.6;// [mole fraction of propane in vapour phase]
+P = 1.5;// [MPa]
+//***************//
+
+// Assume T = 10 OC
+T = 10;// [OC]
+// From Fig. 10.14 (Pg 426):
+K1 = 1.8;
+K2 = 0.5;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+// Since y1 + y2 > 1, so we assume another value of T = 9 OC.
+T = 9;// [OC]
+// From Fig. 10.14 (Pg 426):
+K1 = 1.75;
+K2 = 0.5;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+// Since y1 + y2 = 1. Therefore:
+printf("Bubble Temperature is %d OC\n",T);
+printf("Composition of the vapour bubble:\n y1 = %.2f\n y2 = %.2f",y1,y2);
\ No newline at end of file diff --git a/647/CH10/EX10.15/Example10_15.sce b/647/CH10/EX10.15/Example10_15.sce new file mode 100755 index 000000000..361763b84 --- /dev/null +++ b/647/CH10/EX10.15/Example10_15.sce @@ -0,0 +1,40 @@ +clear;
+clc;
+
+// Example: 10.15
+// Page: 428
+
+printf("Example: 10.15 - Page: 428\n\n");
+
+// Solution
+
+//*****Data******//
+y1 = 0.20;// [mole fraction of methane in vapour phase]
+y2 = 0.30;// [mole fraction of ethane in vapour phase]
+y3 = 0.50;// [mole fraction of propane in vapour phase]
+T = 30;// [OC]
+//*************//
+
+// Assume P = 2.0 MPa
+P = 2.0;// [MPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 8.5;
+K2 = 2.0;
+K3 = 0.68;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 2.15 MPa at 30 OC.
+P = 2.15;// [MPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 8.1;
+K2 = 1.82;
+K3 = 0.62;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Dew Pressure is %.2f MPa\n",P);
+printf("Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f",x1,x2,x3);
\ No newline at end of file diff --git a/647/CH10/EX10.16/Example10_16.sce b/647/CH10/EX10.16/Example10_16.sce new file mode 100755 index 000000000..ee0025c9c --- /dev/null +++ b/647/CH10/EX10.16/Example10_16.sce @@ -0,0 +1,94 @@ +clear;
+clc;
+
+// Example: 10.16
+// Page: 429
+
+printf("Example: 10.16 - Page: 429\n\n");
+
+// Solution
+
+// Dew point Pressure
+//*****Data******//
+y1 = 0.10;// [mole fraction of methane in vapour phase]
+y2 = 0.20;// [mole fraction of ethane in vapour phase]
+y3 = 0.70;// [mole fraction of propane in vapour phase]
+T = 10;// [OC]
+//*************//
+
+// Assume P = 690 kPa
+P = 690;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 20.0;
+K2 = 3.25;
+K3 = 0.92;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 10135 kPa at 10 OC.
+P = 10135;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 13.20;
+K2 = 2.25;
+K3 = 0.65;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 > 1, so we assume another value of P = 870 kPa at 10 OC.
+P = 870;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 16.0;
+K2 = 2.65;
+K3 = 0.762;
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Dew Pressure is %d kPa\n",P);
+printf("Composition of the liquid drop:\n x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n",x1,x2,x3);
+printf("\n");
+
+// Bubble point Pressure
+//*****Data******//
+x1 = 0.10;// [mole fraction of methane in vapour phase]
+x2 = 0.20;// [mole fraction of ethane in vapour phase]
+x3 = 0.70;// [mole fraction of propane in vapour phase]
+T = 10;// [OC]
+//*************//
+
+// Assume P = 2622 kPa
+P = 2622;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.60;
+K2 = 1.11;
+K3 = 0.335;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 > 1, so we assume another value of P = 2760 kPa at 10 OC.
+P = 2760;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.25;
+K2 = 1.07;
+K3 = 0.32;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 < 1, so we assume another value of P = 2656 kPa at 10 OC.
+P = 2656;// [kPa]
+// From Fig. 10.14 (Pg 426):
+K1 = 5.49;
+K2 = 1.10;
+K3 = 0.33;
+// From Eqn. 10.83:
+y1 = K1*x1;
+y2 = K2*x2;
+y3 = K3*x3;
+// Since x1 + x2 +x3 = 1. Therefore:
+printf("Bubble Pressure is %d kPa\n",P);
+printf("Composition of the vapour bubble:\n y1 = %.4f\n y2 = %.4f\n y3 = %.4f",y1,y2,y3);
\ No newline at end of file diff --git a/647/CH10/EX10.17/Example10_17.sce b/647/CH10/EX10.17/Example10_17.sce new file mode 100755 index 000000000..443de93ad --- /dev/null +++ b/647/CH10/EX10.17/Example10_17.sce @@ -0,0 +1,50 @@ +clear;
+clc;
+
+// Example: 10.17
+// Page: 432
+
+printf("Example: 10.17 - Page: 432\n\n");
+
+// Solution
+
+//*****Data******//
+// 1: acetone 2: acetonitrile 3: nitromethane
+z1 = 0.45;
+z2 = 0.35;
+z3 = 0.20;
+P1sat = 195.75;// [kPa]
+P2sat = 97.84;// [kPa]
+P3sat = 50.32;// [kPa]
+//***************//
+
+// Bubble Point Calculation:
+Pbubble = z1*+P1sat + z2*P2sat +z3*P3sat;// [kPa]
+
+// Dew Point Calculation:
+Pdew = 1/((z1/P1sat) + (z2/P2sat) + (z3/P3sat));// [kPa]
+K1 = P1sat/Pdew;
+K2 = P2sat/Pdew;
+K3 = P3sat/Pdew;
+// Overall Material balance:
+// For 1 mol of the feed.
+// L + V = 1......................................... (1)
+// F*zi = L*xi + V*yi ............................... (2)
+// zi = (1 - V)*xi + V*yi ........................... (3)
+// Substituting xi = yi/K in eqn. (3)
+// yi = zi*Ki/(1 + V*(Ki - 1))
+// Since, Sum(yi) = 1.
+deff('[y] = f(V)','y = (z1*K1/(1 + V*(K1 - 1))) + (z2*K2/(1 + V*(K2 - 1))) + (z3*K3/(1 + V*(K3 - 1))) - 1');
+V = fsolve(0.8,f);
+L = 1 - V;
+y1 = z1*K1/(1 + V*(K1 - 1));
+y2 = z2*K2/(1 + V*(K2 - 1));
+y3 = z3*K3/(1 + V*(K3 - 1));
+// From Eqn. 10.83:
+x1 = y1/K1;
+x2 = y2/K2;
+x3 = y3/K3;
+printf(" L = %e mol\n",L);
+printf(" V = %e mol\n",V);
+printf(" y1 = %.4f\n y2 = %.4f\n y3 = %.4f\n",y1,y2,y3);
+printf(" x1 = %.4f\n x2 = %.4f\n x3 = %.4f\n",x1,x2,x3);
\ No newline at end of file diff --git a/647/CH10/EX10.18/Example10_18.sce b/647/CH10/EX10.18/Example10_18.sce new file mode 100755 index 000000000..72db8defc --- /dev/null +++ b/647/CH10/EX10.18/Example10_18.sce @@ -0,0 +1,46 @@ +clear;
+clc;
+
+// Example: 10.18
+// Page: 433
+
+printf("Example: 10.18 - Page: 433\n\n");
+
+// Solution
+
+//*****Data******//
+// 1: Benzene 2: Toulene
+z1 = 0.81;
+Temp = 60;// [OC]
+P = 70;// [kPa]
+// Antonine Constants:
+A1 = 14.2321;
+B1 = 2773.61;
+C1 = 220.13;
+A2 = 15.0198;
+B2 = 3102.64;
+C2 = 220.02;
+//******************//
+
+deff('[P1] = f1(T)','P1 = exp(A1 - B1/(T + C1))');
+P1sat = f1(Temp);// [kPa]
+deff('[P2] = f2(T)','P2 = exp(A2 - B2/(T + C2))');
+P2sat = f2(Temp);// [kPa]
+// P = x1*P1sat + x2*P2sat;
+// x2 = 1 - x1;
+deff('[y] = f3(x1)','[y] = P - (x1*P1sat + (1 - x1)*P2sat)');
+x1 = fsolve(7,f3);
+y1 = x1*P1sat/P;
+x2 = 1 - x1;
+y2 = 1 - y1;
+
+// Basis: 1 mol of feed stream.
+F = 1;// [mol]
+// F*zi = L*xi + V*yi = L*xi + (1 - L)*yi
+deff('[y] = f4(L)','[y] = F*z1 - (L*x1 + (1 - L)*y1)');
+L = fsolve(7,f4);// [mol]
+V = 1 - L;// [mol]
+printf(" L = %.4f mol\n",L);
+printf(" V = %.4f mol\n",V);
+printf(" y1 = %.4f\n y2 = %.4f\n",y1,y2);
+printf(" x1 = %.4f\n x2 = %.4f\n",x1,x2);
\ No newline at end of file diff --git a/647/CH10/EX10.19/Example10_19.sce b/647/CH10/EX10.19/Example10_19.sce new file mode 100755 index 000000000..ecfe4e404 --- /dev/null +++ b/647/CH10/EX10.19/Example10_19.sce @@ -0,0 +1,45 @@ +clear;
+clc;
+
+// Example: 10.19
+// Page: 413
+
+printf("Example: 10.11 - Page: 436\n\n");
+
+// Solution
+
+//*****Data******//
+// (1): acetone (2): carbon tetrachloride
+T = 45;// [OC]
+// Data = [P (torr), x1, y1]
+Data = [315.32 0.0556 0.2165;339.70 0.0903 0.2910;397.77 0.2152 0.4495;422.46 0.2929 0.5137; 448.88 0.3970 0.5832;463.92 0.4769 0.6309;472.84 0.5300 0.6621;485.16 0.6047 0.7081;498.07 0.7128 0.7718;513.20 0.9636 0.9636];
+//*************//
+
+// From the standard data (Pg 531):
+// For Acetone:
+A1 = 14.2342;
+B1 = 2691.46;
+C1 = 230.00;
+// For carbon tetrachloride:
+A2 = 13.6816;
+B2 = 2355.82;
+C2 = 220.58;
+P1sat = exp(A1 - B1/(T + C1));// [kPa]
+P2sat = exp(A2 - B2/(T + C2));// [kPa]
+P1sat = P1sat*760/101.325;// [torr]
+P2sat = P2sat*760/101.325;// [torr]
+P = Data(:,1);
+x1 = Data(:,2);
+y1 = Data(:,3);
+x2 = 1 - x1;
+y2 = 1 - y1;
+gama1 = (y1.*P./x1)/P1sat;
+gama2 = (y2.*P./x2)/P2sat;
+Value = log(gama1./gama2);
+scf(2);
+plot(x1,Value);
+xgrid();
+xlabel("x1");
+ylabel("ln(y1/y2)");
+// Since the whole area is above X - axis:
+printf("The data is not consistent thermodynamically\n");
\ No newline at end of file diff --git a/647/CH10/EX10.2/Example10_2.sce b/647/CH10/EX10.2/Example10_2.sce new file mode 100755 index 000000000..5d26bb565 --- /dev/null +++ b/647/CH10/EX10.2/Example10_2.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 10.2
+// Page: 399
+
+printf("Example: 10.2 - Page: 399\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 10.2 on page number 399 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 10.2 on page 399 of the book.");
\ No newline at end of file diff --git a/647/CH10/EX10.3/Example10_3.sce b/647/CH10/EX10.3/Example10_3.sce new file mode 100755 index 000000000..558d52926 --- /dev/null +++ b/647/CH10/EX10.3/Example10_3.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 10.3
+// Page: 400
+
+printf("Example: 10.3 - Page: 400\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.6;// [mole fraction of ethylene]
+x2 = 0.4;// [mole fraction of propylene]
+T = 423;// [K]
+P1_sat = 15.2;// [vapour pressure of ethylene, atm]
+P2_sat = 9.8;// [vapour pressure of propylene, atm]
+//**************//
+
+P = x1*P1_sat + x2*P2_sat;// [atm]
+printf("The total pressure is %.2f atm\n",P);
+// In vapour phase:
+y1 = x1*P1_sat/P;// [mole fraction of ethylene]
+y2 = x2*P2_sat/P;// [mole fraction of propylene]
+printf("Mole fraction of ethylene in vapour phase is %.1f\n",y1);
+printf("Mole fraction of propylene in the vapour phase is %.1f\n",y2);
\ No newline at end of file diff --git a/647/CH10/EX10.4/Example10_4.sce b/647/CH10/EX10.4/Example10_4.sce new file mode 100755 index 000000000..82a3f8a92 --- /dev/null +++ b/647/CH10/EX10.4/Example10_4.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 10.4
+// Page: 400
+
+printf("Example: 10.4 - Page: 400\n\n");
+
+// Solution
+
+//*****Data******//
+Temp = 77;// [OC]
+P = 75;// [kPa]
+deff('[P1] = f1(T)','P1 = exp(14.35 - 2942/(T + 220))');
+deff('[P2] = f2(T)','P2 = exp(14.25 - 2960/(T + 210))');
+//*************//
+
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+deff('[y] = f3(x1)','y = P - (x1*P1sat) - (1 - x1)*P2sat');
+x1 = fsolve(7,f3);
+x2 = 1 - x1;
+printf("In Liquid phase\n");
+printf("The mole fraction of X is %.3f\n",x1);
+printf("The mole fraction of Y is %.3f\n",x2);
+
+y1 = x1*P1sat/P;
+y2 = 1 - y1;
+printf("In Vapour phase\n");
+printf("The mole fraction of X is %.3f\n",y1);
+printf("The mole fraction of Y is %.3f\n",y2);
\ No newline at end of file diff --git a/647/CH10/EX10.5/Example10_5.sce b/647/CH10/EX10.5/Example10_5.sce new file mode 100755 index 000000000..2cfcf33f2 --- /dev/null +++ b/647/CH10/EX10.5/Example10_5.sce @@ -0,0 +1,58 @@ +clear;
+clc;
+
+// Example: 10.5
+// Page: 401
+
+printf("Example: 10.5 - Page: 401\n\n");
+
+// Solution
+
+//*****Data******//
+deff('[P1] = f1(T)','P1 = exp(14.3916 - 2795/(T + 230))');
+deff('[P2] = f2(T)','P2 = exp(14.2724 - 2945.47/(T + 224))');
+deff('[P3] = f3(T)','P3 = exp(14.2043 - 2972.64/(T + 209))');
+//*************//
+
+// Solution (i)
+
+//*****Data******//
+Temp = 75;// [OC]
+P = 75;// [kPa]
+x1 = 0.30;
+x2 = 0.40;
+//*************//
+
+x3 = 1 - (x1 + x2);
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+P3sat = f3(Temp);// [kPa]
+P = x1*P1sat + x2*P2sat + x3*P3sat;// [kPa]
+y1 = x1*P1sat/P;
+y2 = x2*P2sat/P;
+y3 = x3*P3sat/P;
+printf("Solution (i)\n");
+printf("The mole fraction of acetone is %.3f\n",y1);
+printf("The mole fraction of acetonitrile is %.3f\n",y2);
+printf("The mole fraction of nitromethane is %.3f\n",y3);
+
+// Solution (ii)
+
+//*****Data*****//
+Temp = 80;// [OC]
+y1 = 0.45;
+y2 = 0.35;
+//**************//
+
+y3 = 1 - (y1 + y2);
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+P3sat = f3(Temp);// [kPa]
+P = 1/((y1/P1sat) + (y2/P2sat) + (y3/P3sat));// [kPa]
+x1 = y1*P/P1sat;
+x2 = y2*P/P2sat;
+x3 = y3*P/P3sat;
+printf("Solution (ii)\n");
+printf("The mole fraction of acetone is %.3f\n",x1);
+printf("The mole fraction of acetonitrile is %.3f\n",x2);
+printf("The mole fraction of nitromethane is %.3f\n",x3);
\ No newline at end of file diff --git a/647/CH10/EX10.6/Example10_6.sce b/647/CH10/EX10.6/Example10_6.sce new file mode 100755 index 000000000..4e401790a --- /dev/null +++ b/647/CH10/EX10.6/Example10_6.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 10.6
+// Page: 403
+
+printf("Example: 10.6 - Page: 403\n\n");
+
+// Solution
+
+//*****Data******//
+deff('[P1] = f1(T)','P1 = exp(16.5915 - 3643.31/(T - 33.424))');
+deff('[P2] = f2(T)','P2 = exp(14.2532 - 2665.54/(T - 53.424))');
+deff('[A] = f3(T)','A = 2.771 - 0.00523*T');
+Temp = 318.15;// [K]
+x1 = 0.25;
+//**************//
+
+P1sat = f1(Temp);// [kPa]
+P2sat = f2(Temp);// [kPa]
+A = f3(Temp);
+x2 = 1 - x1;
+gama1 = exp(A*x2^2);
+gama2 = exp(A*x1^2);
+P = x1*gama1*P1sat + x2*gama2*P2sat;
+y1 = x1*gama1*P1sat/P;
+y2 = x2*gama2*P2sat/P;
+printf("In Vapour phase\n");
+printf("The mole fraction of methanol is %.3f\n",y1);
+printf("The mole fraction of methyl acetate is %.3f\n",y2);
\ No newline at end of file diff --git a/647/CH10/EX10.7/Example10_7.sce b/647/CH10/EX10.7/Example10_7.sce new file mode 100755 index 000000000..d61fb697f --- /dev/null +++ b/647/CH10/EX10.7/Example10_7.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Example: 10.7
+// Page: 408
+
+printf("Example: 10.7 - Page: 408\n\n");
+
+// Solution
+
+//*****Data******//
+Temp = 30;// [OC]
+A = 0.625;
+//**************//
+
+P1sat = exp(13.71 - 3800/Temp);// [kPa]
+P2sat = exp(14.01 - 3800/Temp);// [kPa]
+// At azeotropic point:
+// P = gama1*P1sat + gama2*P2sat
+// gama1/gama2 = P2sat/P1sat
+// log(gama1) - log(gama2) = log(P2sat) - log(P1sat)
+// Val = log(gama1) - gama2
+Val = log(P2sat) - log(P1sat);
+// log(gama1) = (A*x2^2)
+// log(gama2) = (A*x1^2)
+// A(x2^2 - x1^2) = 0.625*(x2^2 - x1^2)..................... (1)
+// x1 + x2 = 1............................................. (2)
+// On simplifying, we get:
+// A*(1 - (2*x1)) = Val
+x1 = (1/2)*(1 - Val/A);
+x2 = 1 - x1;
+printf("Azeotropic Composition\n");
+printf("The mole fraction of component 1 is %.3f\n",x1);
+printf("The mole fraction of component 2 is %.3f\n",x2);
\ No newline at end of file diff --git a/647/CH10/EX10.8/Example10_8.sce b/647/CH10/EX10.8/Example10_8.sce new file mode 100755 index 000000000..a3b7043b4 --- /dev/null +++ b/647/CH10/EX10.8/Example10_8.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 10.8
+// Page: 410
+
+printf("Example: 10.8 - Page: 410\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 10.8 on page number 410 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 10.8 on page 410 of the book.");
\ No newline at end of file diff --git a/647/CH10/EX10.9/Example10_9.sce b/647/CH10/EX10.9/Example10_9.sce new file mode 100755 index 000000000..610d9271a --- /dev/null +++ b/647/CH10/EX10.9/Example10_9.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 10.9
+// Page: 412
+
+printf("Example: 10.9 - Page: 412\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.42;
+x2 = 0.58;
+P = 760;// [mm of Hg]
+P1sat = 786;// [mm of Hg]
+P2sat = 551;// [mm of Hg]
+//***************//
+
+gama1 = P/P1sat;
+gama2 = P/P2sat;
+A = log(gama1)*(1 + (x2*log(gama2))/(x1*log(gama1)))^2;
+B = log(gama2)*(1 + (x1*log(gama1))/(x2*log(gama2)))^2;
+printf("Van Laar Constants\n");
+printf("A = %.3f\n",A);
+printf("B = %.3f\n",B);
\ No newline at end of file diff --git a/647/CH11/EX11.1/Example11_1.sce b/647/CH11/EX11.1/Example11_1.sce new file mode 100755 index 000000000..0ea2278e7 --- /dev/null +++ b/647/CH11/EX11.1/Example11_1.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 11.1
+// Page: 458
+
+printf("Example: 11.1 - Page: 458\n\n");
+
+// Solution
+
+//*****Data******//
+a = 2.423;// [g]
+b = 100;// [g]
+Lf = 35.7;// [cal/g]
+Tf = 353.1;// [cal/g]
+delta_Tf = 0.64;// [OC]
+R = 2;// [cal/mol K]
+Mw = 32;// [Molecular wt. of Sulphur, g/mol]
+//*************//
+
+M2 = ((R*Tf^2/(1000*Lf))*(a*1000/(b)))/delta_Tf;// [g/mol]
+n = M2/Mw;
+printf("Molecular Formula of Sulphur is S%d",round(n));
\ No newline at end of file diff --git a/647/CH11/EX11.2/Example11_2.sce b/647/CH11/EX11.2/Example11_2.sce new file mode 100755 index 000000000..0b6d53867 --- /dev/null +++ b/647/CH11/EX11.2/Example11_2.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 11.2
+// Page: 459
+
+printf("Example: 11.2 - Page: 459\n\n");
+
+// Solution
+
+//*****Data******//
+Tf = 5 + 273;// [K]
+Lf = 9830;// [J/mol]
+R = 8.314;// [J/mol K]
+M1 = 78;// [kg/kmol]
+//**************//
+
+Kf = R*Tf^2*M1/(1000*Lf);// [kg/kmol]
+printf("Molal Freezing point is %.2f kg/kmol\n",Kf);
\ No newline at end of file diff --git a/647/CH11/EX11.3/Example11_3.sce b/647/CH11/EX11.3/Example11_3.sce new file mode 100755 index 000000000..5266a572c --- /dev/null +++ b/647/CH11/EX11.3/Example11_3.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 11.3
+// Page: 458
+
+printf("Example: 11.3 - Page: 458\n\n");
+
+// Solution
+
+//*****Data******//
+T_melting = 40;// [OC]
+Tf = T_melting + 273;// [K]
+a = 0.172;// [g]
+b = 12.54;// [g]
+T_new = 39.25;// [OC]
+M2 = 135;// [Molecular wt. of acetanilide, g/mol]
+R = 2;// [cal/mol K]
+//**************//
+
+delta_T = T_melting - T_new;// [OC]
+Kf = delta_T*b*M2/(1000*a);
+Lv = ((R*Tf^2/(1000)))/Kf;// [cal/g]
+printf("Latent Heat of Fusion of phenol is %.2f cal/g\n",Lv);
\ No newline at end of file diff --git a/647/CH11/EX11.4/Example11_4.sce b/647/CH11/EX11.4/Example11_4.sce new file mode 100755 index 000000000..e6fd716aa --- /dev/null +++ b/647/CH11/EX11.4/Example11_4.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 11.4
+// Page: 461
+
+printf("Example: 11.4 - Page: 461\n\n");
+
+// Solution
+
+//*****Data******//
+T_boiling = 118.24;// [OC]
+Tb = T_boiling + 273;// [K]
+a = 0.4344;// [g]
+b = 44.16;// [g]
+Lv = 121;// [cal/g]
+T_new = 118.1;// [OC]
+R = 2;// [cal/mol K]
+//**************//
+
+delta_Tb = T_boiling - T_new;// [OC]
+M2 = (R*Tb^2/(1000*Lv))*(a*1000/(b*delta_Tb));
+printf("Molecular weight of anthracene is %d kg/kmol",round(M2));
\ No newline at end of file diff --git a/647/CH11/EX11.5/Example11_5.sce b/647/CH11/EX11.5/Example11_5.sce new file mode 100755 index 000000000..7aa7c6f63 --- /dev/null +++ b/647/CH11/EX11.5/Example11_5.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 11.5
+// Page: 462
+
+printf("Example: 11.5 - Page: 462\n\n");
+
+// Solution
+
+//*****Data******//
+delta_Tb = 2.3;// [K]
+w1 = 100;// [g]
+M1 = 78;// [g/mol]
+w2 = 13.86;// [g]
+M2 = 154;// [g/mol]
+Tb = 353.1;// [K]
+R = 8.314;// [J/mol K]
+//****************//
+
+// Molality:
+m = w2*1000/(w1*M2);// [mol/kg]
+// Molal Elevation Constant:
+Kb = delta_Tb/m;// [K kg/mol]
+// Molar Latent Heat of Vaporisation:
+Lv = R*Tb^2*M1/(1000*Kb);// [J/mol]
+printf("Molar Latent Heat of Vaporisation is %d J/mol",Lv);
\ No newline at end of file diff --git a/647/CH11/EX11.6/Example11_6.sce b/647/CH11/EX11.6/Example11_6.sce new file mode 100755 index 000000000..eba1bac8d --- /dev/null +++ b/647/CH11/EX11.6/Example11_6.sce @@ -0,0 +1,27 @@ +
+clear;
+clc;
+
+// Example: 11.6
+// Page: 465
+
+printf("Example: 11.6 - Page: 465\n\n");
+
+// Solution
+
+//*****Data******//
+Temp = 50 + 273;// [K]
+w2 = 60;// [g]
+w1 = 1500;// [g]
+M1 = 18;// [g/mol]
+M2 = 180;// [g/mol]
+Vl = 18*10^(-6);// [Molar Volume of water, cubic m/mol]
+R = 8.314;// [J/mol K]
+//***************//
+
+// Mole fraction of glucose:
+x2 = (w2/M2)/((w2/M2) + (w1/M1));
+// Applying Eqn. (11.45):
+P = R*Temp*x2/Vl;// [N/square m]
+P = P/1000;// [kPa]
+printf("Osmotic Pressure is %.2f kPa\n",P);
\ No newline at end of file diff --git a/647/CH11/EX11.7/Example11_7.sce b/647/CH11/EX11.7/Example11_7.sce new file mode 100755 index 000000000..e408ea61e --- /dev/null +++ b/647/CH11/EX11.7/Example11_7.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 11.7
+// Page: 465
+
+printf("Example: 11.7 - Page: 465\n\n");
+
+// Solution
+
+//*****Data******//
+w2 = 0.6;// [g]
+w3 = 1.8;// [g]
+Temp = 27 + 273;// [K]
+V1 = 100;// [cubic cm]
+M2 = 60;// [g/mol]
+M3 = 180;// [g/mol]
+R = 0.082;// [L.atm/mol.K]
+//****************//
+
+V1 = V1/1000;// [litre]
+// C: Concentration per litre
+C = ((w2/M2) + (w3/M3))/V1;// [mol/litre]
+P = C*R*Temp;// [atm]
+printf("Osmotic Pressure of the solution is %.2f atm",P);
\ No newline at end of file diff --git a/647/CH12/EX12.1/Example12_1.sce b/647/CH12/EX12.1/Example12_1.sce new file mode 100755 index 000000000..13a8797f5 --- /dev/null +++ b/647/CH12/EX12.1/Example12_1.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 12.1
+// Page: 471
+
+printf("Example: 12.1 - Page: 471\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 12.1 on page number 471 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 12.1 on page 471 of the book.");
\ No newline at end of file diff --git a/647/CH12/EX12.10/Example12_10.sce b/647/CH12/EX12.10/Example12_10.sce new file mode 100755 index 000000000..3c1d4c9fa --- /dev/null +++ b/647/CH12/EX12.10/Example12_10.sce @@ -0,0 +1,40 @@ +clear;
+clc;
+
+// Example: 12.10
+// Page: 489
+
+printf("Example: 12.10 - Page: 489\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: CO(g) + H2O(g) ------> CO2(g) + H2(g)
+P = 10;// [bar]
+T = 1000;// [K]
+K_1000 = 1.5;
+//***********//
+
+// Moles in feed:
+nCO_feed = 1;
+nH20_feed = 1;
+// Let e be the degree of completion at equilibrium.
+// Moles at Equilibrium:
+// nCO_eqb = 1 - e;
+// nH20_eqb = 1 - e;
+// nCO2_eqb = e;
+// nH2_eqb = e;
+// Total moles at equilibrium = 1 - e + 1 - e + e + e = 2
+// Mole Fractions at Equilibrium:
+// yCO_eqb = (1 - e)/2;
+// yH20_eqb = (1 - e)/2;
+// yCO2_eqb = e/2;
+// yH2_eqb = e/2;
+// Sum of stoichometric coeffecient:
+v = 1 + 1 - 1 - 1;
+K = K_1000*P^v;
+deff('[y] = f(e)','y = K - (e/2)*(e/2)/(((1 - e)/2)*(1 - e)/2)');
+e = fsolve(0.5,f);
+printf("Composition of the gas leaving the mixture\n");
+printf("yCO = yH20 = %.4f\n",(1 - e)/2);
+printf("yCO2 = yH2 = %.4f\n",e/2);
\ No newline at end of file diff --git a/647/CH12/EX12.11/Example12_11.sce b/647/CH12/EX12.11/Example12_11.sce new file mode 100755 index 000000000..e7424d731 --- /dev/null +++ b/647/CH12/EX12.11/Example12_11.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+// Example: 12.11
+// Page: 489
+
+printf("Example: 12.11 - Page: 489\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: N2(g) + 3H2(g) --------> 2NH3
+nN2_feed = 1;
+nH2_feed = 5;
+T = 800;// [K]
+P = 250;// [bar]
+K = 1.106*10^(-5);
+//**************//
+
+// Let e be the degree of completion at equilibrium.
+// Moles at Equilibrium:
+// nN2_eqb = 1 - e;
+// nH2_eqb = 5 - 3*e;
+// nNH3_eqb = 2*e;
+// Total moles at equilibrium = 1 - e + 5 - 3*e + 2*e = 2*(3 - e)
+// Mole Fractions at Equilibrium:
+// yN2_eqb = (1 - e)/(2*(3 - e));
+// yH2_eqb = (5 - 3*e)/(2*(3 - e));
+// yNH3_eqb = 2*e/(2*(3 - e));
+// Sum of stoichometric coeffecient:
+v = 2 - 3 - 1;
+Ky = K*P^(-v);
+deff('[y] = f(e)','y = Ky - ((2*e/(2*(3 - e)))^2)/(((1-e)/((2*(3 - e))))*((5 - 3*e)/(2*(3 - e)))^3)');
+e = fsolve(0.5,f);
+printf("Molar Composition of the gases\n");
+printf("yN2 = %.4f\n",(1 - e)/(2*(3 - e)));
+printf("yH2 = %.4f\n",(5 - 3*e)/(2*(3 - e)));
+printf("yNH3 = %.4f\n",(2*e)/(2*(3 - e)));
\ No newline at end of file diff --git a/647/CH12/EX12.12/Example12_12.sce b/647/CH12/EX12.12/Example12_12.sce new file mode 100755 index 000000000..d04ad310d --- /dev/null +++ b/647/CH12/EX12.12/Example12_12.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+
+// Example: 12.12
+// Page: 490
+
+printf("Example: 12.12 - Page: 490\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: SO2(g) + (1/2)O2 ------> SO3(g)
+P = 1;// [bar]
+T = 750;// [K]
+K = 74;
+//************//
+
+// Moles in Feed:
+nSO2 = 1;
+nO2 = 0.5;
+// Let e be the degree of completion at equilibrium.
+// Moles at Equilibrium:
+// nSO2_eqb = 10 - e;
+// nO2_eqb = 8 - 0.5*e;
+// nSO3_eqb = e;
+// Total no. of moles = 10 - e + 8 - 0.5*e + e = 18 -0.5*e;
+// Mole fraction at Equilibrium:
+// ySO2_eqb = (10 - e)/(18 - 0.5*e);
+// yO2_eqb = (8 - 0.5*e)/(18 - 0.5*e);
+// ySO3_eqb = e/(18 - 0.8*e);
+// Sum of stoichometric coeffecient:
+v = 1 - 1 -0.5;
+Ky = K*P^(-v);
+deff('[y] = f(e)','y = Ky - (e*(18 - (0.5*e)))/((10 - e)*(8 - 0.5*e))');
+e = fsolve(7,f);
+printf("Molar Composition of the gases\n");
+printf("nSO2 = %.2f\n",(10 - e));
+printf("nO2 = %.2f\n",(8 - 0.5*e));
+printf("nSO3 = %.2f\n",e);
\ No newline at end of file diff --git a/647/CH12/EX12.13/Example12_13.sce b/647/CH12/EX12.13/Example12_13.sce new file mode 100755 index 000000000..bebafb163 --- /dev/null +++ b/647/CH12/EX12.13/Example12_13.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 12.13
+// Page: 492
+
+printf("Example: 12.13 - Page: 492\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: PCl5 = PCl3 + Cl2
+T = 250;// [OC]
+Kp = 1.8;
+e = 0.5;
+//**************//
+
+// Basis: 1 mol of PCl5
+// At Equilibrium:
+n_PCl5 = 1 - e;
+n_PCl3 = e;
+n_Cl2 = e;
+n_total = n_PCl5 + n_PCl3 + n_Cl2;
+// Patrial Pressures:
+// P_PCl5 = (n_PCl5/n_total)*P
+// P_PCl3 = (n_PCl3/n_total)*P
+// P_Cl2 = (n_Cl2/n_total)*P
+deff('[y] = f(P)','y = Kp - ((n_PCl3/n_total)*P)*((n_Cl2/n_total)*P)/((n_PCl5/n_total)*P)');
+P = fsolve(7,f);// [atm]
+printf("Total Pressure Required for 50 %% conversion of PCl5 is %.1f atm",P);
\ No newline at end of file diff --git a/647/CH12/EX12.14/Example12_14.sce b/647/CH12/EX12.14/Example12_14.sce new file mode 100755 index 000000000..ee66f0525 --- /dev/null +++ b/647/CH12/EX12.14/Example12_14.sce @@ -0,0 +1,53 @@ +clear;
+clc;
+
+// Example: 12.14
+// Page: 494
+
+printf("Example: 12.14 - Page: 494\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: SO2(g) + (1/2)O2(g) -------> SO3(g)
+// Moles in Feed:
+nSO2_feed = 1;
+nO2_feed = 0.5;
+nAr_feed = 2;
+P = 30;// [bar]
+T = 900;// [K]
+K = 6;
+//*************//
+
+// Let e be the degree of completion at equilibrium.
+// nSO2_eqb = 1 - e;
+// nO2_eqb = 0.5*(1 - e);
+// nSO3_eqb = e;
+// nAr_eqb = 2;
+// Total moles at equilibrium = 1 - e + 0.5*(1 - e) + e + 2 = (7 - e)/2
+// Mole fractions:
+// ySO2_eqb = 2*(1 - e)/(7 - e)
+// yO2_eqb = (1 - e)/(7 - e)
+// ySO3_eqb = 2*e/(7 - e)
+// yAr_eqb = 4/(7 - e)
+// Sum of Stoichiometric Coeffecient:
+v = 1 - 1 - 1/2;
+Ky = K*P^(-v);
+e = 0.8;
+err = 1;
+while err > 0.2
+ Ky_new = (2*e/(7 - e))/(((2*(1 - e))/(7 - e))*(((1 - e)/(7 - e))^0.5));
+ err = abs(Ky - Ky_new);
+ Ky = Ky_new;
+ e = e + 0.001;
+end
+printf("Degree of Conversion is %.3f\n",e);
+ySO2_eqb = 2*(1 - e)/(7 - e);
+yO2_eqb = (1 - e)/(7 - e);
+ySO3_eqb = 2*e/(7 - e);
+yAr_eqb = 4/(7 - e);
+printf("Equilibrium Composition of the reaction Mixture\n");
+printf("ySO2 = %.4f\n",ySO2_eqb);
+printf("yO2 = %.4f\n",yO2_eqb);
+printf("ySO3 = %.4f\n",ySO3_eqb);
+printf("yAr = %.4f\n",yAr_eqb)
\ No newline at end of file diff --git a/647/CH12/EX12.15/Example12_15.sce b/647/CH12/EX12.15/Example12_15.sce new file mode 100755 index 000000000..903734527 --- /dev/null +++ b/647/CH12/EX12.15/Example12_15.sce @@ -0,0 +1,47 @@ +clear;
+clc;
+
+// Example: 12.15
+// Page: 498
+
+printf("Example: 12.15 - Page: 498\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: CH3COOH(l) + C2H5OH(l) --------> CH3COOC2H5(l) + H2O(l)
+T = 373.15;// [K]
+nCH3COOH_feed = 1;
+nC2H5OH_feed = 1;
+deltaHf_CH3COOH = -484.5;// [kJ]
+deltaHf_C2H5OH = -277.69;// [kJ]
+deltaHf_CH3COOC2H5 = -480;// [kJ]
+deltaHf_H2O = -285.83;// [kJ]
+deltaGf_CH3COOH = -389.9;// [kJ]
+deltaGf_C2H5OH = -174.78;// [kJ]
+deltaGf_CH3COOC2H5 = -332.2;// [kJ]
+deltaGf_H2O = -237.13;// [kJ]
+R = 8.314;// [J/mol K]
+//******************//
+
+deltaH_298 = deltaHf_CH3COOC2H5 + deltaHf_H2O - deltaHf_CH3COOH - deltaHf_C2H5OH;// [kJ]
+deltaG_298 = deltaGf_CH3COOC2H5 + deltaGf_H2O - deltaGf_CH3COOH - deltaGf_C2H5OH;// [kJ]
+T0 = 298;// [K]
+K_298 = exp(-(deltaG_298*1000/(R*T0)));
+K_373 = K_298*exp((deltaH_298*1000/R)*((1/T0) - (1/T)));
+// Let e be the degree of completion at equilibrium.
+// nCH3COOH_eqb = 1 - e;
+// nC2H5OH_eqb = 1 - e;
+// nCH3COOC2H5_eqb = e;
+// nH2O_eqb = e;
+// Total moles at equilibrium = 1 - e + 1 - e + e + e = 2
+// Mole fractions:
+// ySO2_eqb = (1 - e)/2
+// yO2_eqb = (1 - e)/2
+// ySO3_eqb = e/2
+// yAr_eqb = e/2
+// Sum of Stoichiometric Coeffecient:
+v = 1 + 1 - 1 - 1;
+deff('[y] = f(e)','y = K_373 - ((e/2)*(e/2))/(((1 - e)/2)*((1 - e)/2))');
+e = fsolve(0.5,f);
+printf("Mole fraction of ethyl acetate is %.3f",e/2);
\ No newline at end of file diff --git a/647/CH12/EX12.16/Example12_16.sce b/647/CH12/EX12.16/Example12_16.sce new file mode 100755 index 000000000..3bb7463ad --- /dev/null +++ b/647/CH12/EX12.16/Example12_16.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 12.16
+// Page: 501
+
+printf("Example: 12.16 - Page: 501\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: CaCO3(s) = CaO(s) + CO2(g)
+T = 1000;// [K]
+deltaH_1000 = 1.7533*10^5;// [J]
+deltaS_1000 = 150.3;// [J/mol K]
+R = 8.314;// [J/mol K]
+//****************//
+
+deltaG_1000 = deltaH_1000 - T*deltaS_1000;// [J]
+K_1000 = exp(-(deltaG_1000/(R*T)));// [bar]
+P = K_1000;
+printf("The decomposition pressure of limestone is %.4f bar at 1000 K",P);
\ No newline at end of file diff --git a/647/CH12/EX12.17/Example12_17.sce b/647/CH12/EX12.17/Example12_17.sce new file mode 100755 index 000000000..5988e86e3 --- /dev/null +++ b/647/CH12/EX12.17/Example12_17.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 12.17
+// Page: 501
+
+printf("Example: 12.17 - Page: 501\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: A(s) ---------> B(s) + C(g)
+deff('[deltaG] = f1(T)','deltaG = 85000 - 213.73*T + 6.71*T*log(T) - 0.00028*T^2');// [J]
+T1 = 400;// [K]
+T2 = 700;// [K]
+Pc = 1;// [bar]
+R = 8.314;// [J/mol K]
+//**************//
+
+deltaG_400 = f1(400);// [J]
+deltaG_700 = f1(700);// [J]
+K_400 = exp(-(deltaG_400/(R*T1)));// [bar]
+K_700 = exp(-(deltaG_700/(R*T2)));// [bar]
+printf("The decomposition pressure is %.4f bar at 400 K\n",K_400);
+printf("The decomposition pressure is %.2f bar at 700 K\n",K_700);
+
+// Equilibrium constant for solid - gas reaction is:
+// K = aB*aC/aA = aC = fC = Pc
+deff('[y] = f2(T)','y = Pc - exp(-f1(T)/(R*T))');
+T = fsolve(900,f2);// [K]
+printf("The decomposition temperature is %.3f K",T);
\ No newline at end of file diff --git a/647/CH12/EX12.18/Example12_18.sce b/647/CH12/EX12.18/Example12_18.sce new file mode 100755 index 000000000..4c1dea6fb --- /dev/null +++ b/647/CH12/EX12.18/Example12_18.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+// Example: 12.18
+// Page: 502
+
+printf("Example: 12.18 - Page: 502\n\n");
+
+// Solution
+
+//*****Data******//
+T = 875;// [K]
+K = 0.514;
+P = 1;// [bar]
+//*************//
+
+// Reaction: Fe(s) + H2O(g) --------->FeO(s) + H2(g)
+// K = a_FeO*a_H2/(a_Fe*a_H2O)
+// Since the activities of the solid components Fe & FeO at equilibrium may be taken as unity.
+// a_Fe = a_FeO = 1
+// Ka = a_H2/a_H2O;
+// Feed:
+nH2O_feed = 1;
+nH2feed = 0;
+// Let e be the degree of completion at equilibrium.
+// Moles at Equilibrium:
+// nH2O_eqb = 1 - e;
+// nH2_eqb = e;
+// Total moles at equilibrium = 1 - e + e = 1
+// Mole Fractions at Equilibrium:
+// yH20_eqb = 1 - e;
+// yH2_eqb = e;
+// Sum of stoichometric coeffecient:
+v = 1 - 1;
+Ky = K*P^(-v);
+// Ky = e/(1 - e)
+e = Ky/(Ky + 1);
+yH2_eqb = e;
+yH2O_eqb = 1 - e;
+printf("Equilibrium Composition\n");
+printf("yH2 = %.2f\n",yH2_eqb);
+printf("y_H2O = %.2f\n",yH2O_eqb);
\ No newline at end of file diff --git a/647/CH12/EX12.19/Example12_19.sce b/647/CH12/EX12.19/Example12_19.sce new file mode 100755 index 000000000..cc82d408d --- /dev/null +++ b/647/CH12/EX12.19/Example12_19.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 12.19
+// Page: 503
+
+printf("Example: 12.19 - Page: 503\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 12.19 on page number 503 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 12.19 on page 503 of the book.");
\ No newline at end of file diff --git a/647/CH12/EX12.2/Example12_2.sce b/647/CH12/EX12.2/Example12_2.sce new file mode 100755 index 000000000..773766121 --- /dev/null +++ b/647/CH12/EX12.2/Example12_2.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 12.2
+// Page: 473
+
+printf("Example: 12.2 - Page: 473\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 12.2 on page number 473 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 12.2 on page 473 of the book.");
\ No newline at end of file diff --git a/647/CH12/EX12.20/Example12_20.sce b/647/CH12/EX12.20/Example12_20.sce new file mode 100755 index 000000000..ae651ae35 --- /dev/null +++ b/647/CH12/EX12.20/Example12_20.sce @@ -0,0 +1,37 @@ +clear;
+clc;
+
+// Example: 12.20
+// Page: 505
+
+printf("Example: 12.20 - Page: 505\n\n");
+
+// Solution
+
+//Reactions:
+// CO + (1/2)O2 ------------> CO2 ......................................(1)
+// C + O2 ------------------> CO2 ......................................(2)
+// H2 + (1/2)O2 ------------> H2O ......................................(3)
+// C + 2H2 -----------------> CH4 ......................................(4)
+
+// Elimination of C:
+// Combining Eqn. (2) with (1):
+// CO + (1/2)O2 ------------> CO2 ......................................(5)
+// Combining Eqn. (2) with (4):
+// CH4 + O2 ----------------> 2H2 + CO2 ................................(6)
+
+// Elimination of O2:
+// Combining Eqn. (3) with (4):
+// CO2 + H2 ----------------> CO + H2O .................................(7)
+// Combining Eqn. (3) with (6):
+// CH4 + 2H2O -------------> CO2 + 4H2 .................................(8)
+
+// Equations 7 & 8 are independent sets. Hence
+r = 2;// [No. of independent rkn.]
+C = 5;// [No. of component]
+P = 1;// [No. of phases]
+s = 0;// [No special constraint]
+// Applying Eqn. 12.81
+F = C - P + 2 - r - s;// [Degree of freedom]
+printf("No. of independent reaction that occur is %d\n",r);
+printf("No. of Degree of freedom is %d",F);
\ No newline at end of file diff --git a/647/CH12/EX12.21/Example12_21.sce b/647/CH12/EX12.21/Example12_21.sce new file mode 100755 index 000000000..ad564f631 --- /dev/null +++ b/647/CH12/EX12.21/Example12_21.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Example: 12.21
+// Page: 506
+
+printf("Example: 12.21 - Page: 506\n\n");
+
+// Solution
+
+// Reaction: CaCO3 -----------> CaO + CO2
+r = 1;// [No. of independent rkn.]
+C = 3;// [No. of component]
+P = 3;// [No. of phases, solid CaO, solid CaCO3, gaseous CO2]
+s = 0;// [No special constraint]
+// Applying Eqn. 12.81
+F = C - P + 2 - r - s;// [Degree of freedom]
+printf("No. of Degree of freedom is %d",F);
\ No newline at end of file diff --git a/647/CH12/EX12.22/Example12_22.sce b/647/CH12/EX12.22/Example12_22.sce new file mode 100755 index 000000000..2a14fa441 --- /dev/null +++ b/647/CH12/EX12.22/Example12_22.sce @@ -0,0 +1,76 @@ +clear;
+clc;
+
+// Example: 12.22
+// Page: 508
+
+printf("Example: 12.22 - Page: 508\n\n");
+
+// Solution
+
+//*********Data*********//
+// Reaction:
+// C4H10 -----------> C2H4 + C2H6 ....................................(A)
+// C4H10 -----------> C3H6 + CH4 .................................... (B)
+T = 750;// [K]
+P = 1.2;// [bar]
+Ka = 3.856;
+Kb = 268.4;
+//************************//
+
+// Let
+// ea = Degree of conversion of C4H10 in reaction (A)
+// eb = Degree of conversion of C4H10 in reaction (B)
+
+// Moles in Feed:
+nC4H10_feed = 1;
+nC2H4_feed = 0;
+nC2H6_feed = 0;
+nC3H6_feed = 0;
+nCH4_feed = 0;
+
+// Moles at Equilibrium:
+// nC4H10_eqb = 1 - ea - eb
+// nC2H4_eqb = ea
+// nC2H6_eqb = ea
+// nC3H6_eqb = eb
+// nCH4_eqb = eb
+
+// Total moles at equilibrium = 1 - ea - eb + ea + eb + eb = 1 + ea + eb
+
+// Mole Fraction:
+// yC4H10_eqb = (1 - ea - eb)/(1 + ea + eb)
+// yC2H4_eqb = ea/(1 + ea + eb)
+// yC2H6_eqb = ea/(1 + ea + eb)
+// yC3H6_eqb = eb/(1 + ea + eb)
+// yCH4_eqb = eb/(1 + ea + eb)
+
+// Sum of the stoichometric coeffecient:
+va = 1 + 1 - 1;
+vb = 1 + 1 - 1;
+
+// e = [ea eb]
+// Solution of simultaneous equation
+function[f]=F(e)
+ f(1) = (((e(1)/(1 + e(1) + e(2)))*(e(1)/(1 + e(1) + e(2))))/((1 - e(1) - e(2))/(1 + e(1) + e(2))))*P^va - Ka;
+ f(2) = (((e(2)/(1 + e(1) + e(2)))*(e(2)/(1 + e(1) + e(2))))/((1 - e(1) - e(2))/(1 + e(1) + e(2))))*P^vb - Kb;
+ funcprot(0);
+endfunction
+
+// Initial guess:
+e = [0.1 0.8];
+y = fsolve(e,F);
+ea = y(1);
+eb = y(2);
+yC4H10_eqb = (1 - ea - eb)/(1 + ea + eb);
+yC2H4_eqb = ea/(1 + ea + eb);
+yC2H6_eqb = ea/(1 + ea + eb);
+yC3H6_eqb = eb/(1 + ea + eb);
+yCH4_eqb = eb/(1 + ea + eb);
+
+printf("At Equilibrium\n");
+printf("yC4H10 = %.4f\n",yC4H10_eqb);
+printf("yC2H4 = %.4f\n",yC2H4_eqb);
+printf("yC2H6 = %.4f\n",yC2H6_eqb);
+printf("yC3H6 = %.4f\n",yC3H6_eqb);
+printf("yCH4 = %.4f\n",yCH4_eqb);
\ No newline at end of file diff --git a/647/CH12/EX12.3/Example12_3.sce b/647/CH12/EX12.3/Example12_3.sce new file mode 100755 index 000000000..435338b00 --- /dev/null +++ b/647/CH12/EX12.3/Example12_3.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+// Example: 12.3
+// Page: 479
+
+printf("Example: 12.3 - Page: 479\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: C2H5OH(g) + (1/2)O2(g) = CH3CHO(g) + H2O(g)
+Temp = 298;// [K]
+G_CH3CHO = -133.978;// [kJ]
+G_H2O = -228.60;// [kJ]
+G_C2H5OH = -174.883;// [kJ]
+R = 8.314;// [J/mol K]
+//***************//
+
+G_O2 = 0;// [kJ]
+G_rkn = G_CH3CHO + G_H2O -(G_C2H5OH + G_O2);// [kJ]
+G_rkn = G_rkn*1000;// [J]
+if G_rkn < 0
+ printf("Reaction is feasible\n");
+ K = exp(-(G_rkn/(R*Temp)));
+ printf("Equilibium Constant is %.3e",K);
+else
+ printf("Reaction is not feasible\n");
+end
\ No newline at end of file diff --git a/647/CH12/EX12.4/Example12_4.sce b/647/CH12/EX12.4/Example12_4.sce new file mode 100755 index 000000000..bfaa2cfca --- /dev/null +++ b/647/CH12/EX12.4/Example12_4.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 12.4
+// Page: 479
+
+printf("Example: 12.4 - Page: 479\n\n");
+
+// Solution
+
+//*****Data******//
+// H2O(l) = H2O(g)
+deltaH = 9710;// [cal]
+deltaS = 26;// [e.u.]
+Temp = 27 + 273;// [K]
+P = 1;// [atm]
+//**************//
+
+deltaG = deltaH - Temp*deltaS;// [cal]
+if deltaG > 0
+ printf("Vaporisation of liquid water is not spontaneous\n");
+else
+ printf("Vaporisation of liquid water is spontaneous")
+end
\ No newline at end of file diff --git a/647/CH12/EX12.5/Example12_5.sce b/647/CH12/EX12.5/Example12_5.sce new file mode 100755 index 000000000..e5435c7de --- /dev/null +++ b/647/CH12/EX12.5/Example12_5.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 12.5
+// Page: 481
+
+printf("Example: 12.5 - Page: 481\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: N2(g) + 3H2(g) = 2NH3(g)
+Temp = 298;// [K]
+G_NH3 = -16.750;// [kJ/mol]
+R = 8.314;// [J/mol K]
+//***************//
+
+G_N2 = 0;// [kJ/mol]
+G_H2 = 0;// [kJ/mol]
+G_rkn = 2*G_NH3 - G_N2 - 3*G_H2;// [kJ/mol]
+printf("Standard Gibbs free energy change is %.1f kJ/mol\n",G_rkn);
+G_rkn = G_rkn*1000;// [J/mol];
+K = exp(-(G_rkn/(R*Temp)));
+printf("Equilibrium constant is %.2e",K);
\ No newline at end of file diff --git a/647/CH12/EX12.6/Example12_6.sce b/647/CH12/EX12.6/Example12_6.sce new file mode 100755 index 000000000..cb95afa2f --- /dev/null +++ b/647/CH12/EX12.6/Example12_6.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 12.6
+// Page: 481
+
+printf("Example: 12.6 - Page: 481\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: CO(g) + H2O(g) = CO2(g) + H2(g)
+G_CO = -32.8;// [kcal]
+G_H2O = -54.64;// [kcal]
+G_CO2 = -94.26;// [kcal]
+Temp = 273 + 25;// [K]
+R = 1.987;// [cal/mol.K]
+//***************//
+
+G_H2 = 0;// [kcal]
+G_rkn = G_CO2 + G_H2 - (G_CO + G_H2O);// [kcal]
+G_rkn = G_rkn*1000;// [cal]
+K = exp(-(G_rkn/(R*Temp)));
+printf("Equilibrium Constant is %.3e",K);
\ No newline at end of file diff --git a/647/CH12/EX12.7/Example12_7.sce b/647/CH12/EX12.7/Example12_7.sce new file mode 100755 index 000000000..51916217e --- /dev/null +++ b/647/CH12/EX12.7/Example12_7.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 12.7
+// Page: 485
+
+printf("Example: 12.7 - Page: 485\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: CO(g) + H2O(g) = CO2(g) + H2(g)
+T1 = 298;// [K]
+T2 = 1000;// [K]
+P = 1;// [bar]
+K1 = 1.1582*10^5;
+H_CO = -110.532;// [kJ]
+H_H2O = -241.997;// [kJ]
+H_CO2 = -393.978;// [kJ]
+R = 8.314;// [J/mol.K]
+//***************//
+
+H_H2 = 0;// [kJ]
+H_rkn = H_CO2 + H_H2 - (H_CO + H_H2O);// [kJ]
+H_rkn = H_rkn*1000;// [J]
+// From Van't Hoff Equation,
+K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)));
+printf("Equilibrium constant at 1000 K is %.4f",K2);
\ No newline at end of file diff --git a/647/CH12/EX12.8/Example12_8.sce b/647/CH12/EX12.8/Example12_8.sce new file mode 100755 index 000000000..614127b0c --- /dev/null +++ b/647/CH12/EX12.8/Example12_8.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 12.8
+// Page: 485
+
+printf("Example: 12.8 - Page: 485\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: N2(g) + 3H2(g) = 2NH3(g)
+T1 = 298;// [K]
+T2 = 700;// [K]
+H_NH3 = -46.1;// [kJ]
+G_NH3 = -16.747;// [kJ]
+R = 8.314;// [J/mol.K]
+//**************//
+
+H_N2 = 0;// [kJ]
+H_H2 = 0;// [kJ]
+G_N2 = 0;// [kJ]
+G_H2 = 0;// [kJ]
+H_rkn = 2*H_NH3 - (H_N2 + 3*H_H2);// [kJ]
+G_rkn = 2*G_NH3 - (G_N2 + 3*G_H2);// [kJ]
+H_rkn = H_rkn*1000;// [J]
+G_rkn = G_rkn*1000;// [J]
+K1 = exp(-(G_rkn/(R*T1)));
+K2 = K1*exp((H_rkn/R)*((1/T1) - (1/T2)));
+printf("Equilibrium constant at 700 K is %.4e",K2);
\ No newline at end of file diff --git a/647/CH12/EX12.9/Example12_9.sce b/647/CH12/EX12.9/Example12_9.sce new file mode 100755 index 000000000..0cdd2d1e7 --- /dev/null +++ b/647/CH12/EX12.9/Example12_9.sce @@ -0,0 +1,40 @@ +clear;
+clc;
+
+// Example: 12.9
+// Page: 486
+
+printf("Example: 12.9 - Page: 486\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: CO(g) + H2O(g) ----> CO2(g) + H2(g)
+T1 = 298;// [K]
+T2 = 1000;// [K]
+deltaH_298 = -41450;// [J/mol]
+deltaGf_298 = -28888;// [J/mol]
+alpha_CO2 = 45.369;// [kJ/kmol K]
+alpha_H2 = 27.012;// [kJ/kmol K]
+alpha_CO = 28.068;// [kJ/kmol K]
+alpha_H2O = 28.850;// [kJ/kmol K]
+beeta_CO2 = 8.688*10^(-3);// [kJ/kmol square K]
+beeta_H2 = 3.509*10^(-3);// [kJ/kmol square K]
+beeta_CO = 4.631*10^(-3);// [kJ/kmol square K]
+beeta_H2O = 12.055*10^(-3);// [kJ/kmol square K]
+R = 8.314;// [J/mol K]
+//*************//
+
+delta_alpha = alpha_CO2 + alpha_H2 - (alpha_CO + alpha_H2O);
+delta_beeta = beeta_CO2 + beeta_H2 - (beeta_CO + beeta_H2O);
+// To obtain the standard heat of reaction:
+deltaH_0 = deltaH_298 - (delta_alpha*T1 + (delta_beeta*T1^2)/2);// [kJ/mol]
+// From Eqn. 12.52:
+IR = (deltaH_0 - delta_alpha*T1*log(T1) - (delta_beeta*T1^2)/2 - deltaGf_298)/T1;// [kJ/mol K]
+// Substituting T = T2 and IR in Eqn. 12.51:
+deltaG_1000 = deltaH_0 - delta_alpha*T2*log(T2) - (delta_beeta*T2^2)/2 - IR*T2;// [kJ/mol]
+printf("Standard Gibbs free energy at 1000 K %.3f kJ\n",deltaG_1000/1000);
+
+// Standard Equilibrium Constant at 1000 K
+K_1000 = exp(-(deltaG_1000)/(R*T2));
+printf("Standard Equilibrium Constant is %.1f",K_1000);
\ No newline at end of file diff --git a/647/CH2/EX2.1/Example2_1.sce b/647/CH2/EX2.1/Example2_1.sce new file mode 100755 index 000000000..aaff9dbb9 --- /dev/null +++ b/647/CH2/EX2.1/Example2_1.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 2.1
+// Page: 39
+
+printf("Example: 2.1 - Page: 39\n\n");
+
+// Solution
+
+//*****Data*****//
+deff('[E] = f1(T)','E = 50 + 25*T + 0.05*T^2');// [J]
+deff('[Q] = f2(T)','Q = 4000 + 10*T');// [J]
+Ti = 400;// [K]
+Tf = 800;// [K]
+//*************//
+
+// From the first law of thermodynamics:
+// W = Q - delta_E
+// W = f2 -f1
+W = integrate('(4000 + 10*T) - (50 + (25*T) + (0.05*T^2))','T',Ti,Tf);
+printf("The work done during the process is %.2f kJ\n",W/1000);
\ No newline at end of file diff --git a/647/CH2/EX2.10/Example2_10.sce b/647/CH2/EX2.10/Example2_10.sce new file mode 100755 index 000000000..c533cb897 --- /dev/null +++ b/647/CH2/EX2.10/Example2_10.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 2.10
+// Page: 53
+
+printf("Example: 2.10 - Page: 53\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 300;// [K]
+V1 = 30;// [L]
+V2 = 3;// [L]
+Cv = 5;// [cal/mol]
+R = 2;// [cal/K mol]
+//*************//
+
+Cp = Cv + R;// [cal/mol]
+gama = Cp/Cv;
+// The relation between temperature and volume of ideal gas undergoing adiabatic change is given by:
+// (T2/T1) = (V1/V2)^(gama - 1)
+T2 = T1 * (V1/V2)^(gama - 1);// [K]
+printf("The final temperature is %.1f K\n",T2);
\ No newline at end of file diff --git a/647/CH2/EX2.11/Example2_11.sce b/647/CH2/EX2.11/Example2_11.sce new file mode 100755 index 000000000..b86aef6f4 --- /dev/null +++ b/647/CH2/EX2.11/Example2_11.sce @@ -0,0 +1,37 @@ +clear;
+clc;
+
+// Example: 2.11
+// Page: 53
+
+printf("Example: 2.11 - Page: 53\n\n");
+
+// Solution
+
+//*****Data*****//
+n = 2;// [mol]
+T1 = 293;// [K]
+P1 = 15;//[atm]
+P2 = 5;// [atm]
+Cp = 8.58;// [cal/degree mol]
+//**************//
+
+R = 2;// [cal/degree mol]
+Cv = Cp - R;// [cal /degree mol]
+gama = Cp/Cv;
+R = 0.082;// [L atm/degree K]
+// Since the gas is ideal:
+V1 = n*R*T1/P1;// [L]
+// Under adiabatic conditions:
+// (V2/V1) = (P1/P2)^(1/gama)
+V2 = V1*(P1/P2)^(1/gama);// [L]
+printf("The final volme is %.2f L\n",V2);
+
+// To determine the final temperature:
+// (T2/T1) = (V1/V2)^(gama - 1);
+T2 = T1*(V1/V2)^(gama - 1);// [K]
+printf("The final temperature is %.2f K\n",T2);
+
+// Adiabatic Work done can be calculated as:
+W = (P1*V1 - P2*V2)/(gama - 1);
+printf("Adiabatic work done is %.2f L-atm\n",W);
\ No newline at end of file diff --git a/647/CH2/EX2.12/Example2_12.sce b/647/CH2/EX2.12/Example2_12.sce new file mode 100755 index 000000000..7b3e2ee39 --- /dev/null +++ b/647/CH2/EX2.12/Example2_12.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+// Example: 2.12
+// Page: 57
+
+printf("Example: 2.12 - Page: 57\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 1;// [kg]
+P1 = 8;// [atm]
+T1 = 50 + 273;// [K]
+// V1 = V;// [L]
+// V2 = 5V;// [L]
+V1_by_V2 = 1/5;
+gama = 1.4;
+R = 0.082;// [L-atm]
+//***************//
+
+// Adiabatic process:
+printf("Adiabatic Process \n");
+P2 = P1*V1_by_V2^gama;// [atm]
+printf("Final Pressure is %.2f atm\n",P2);
+T2 = T1*V1_by_V2^(gama - 1);// [K]
+printf("Final Temperature is %f K\n",T2);
+Wad = R*(T2 - T1)/(1 - gama);// [L-atm]
+printf("Adiabatic Work done is %.3f L-atm\n",Wad);
+printf("\n")
+
+// Isothermal Process:
+printf("Isothermal Process\n")
+// In an isothermal Process, the temperature remans constant:
+T2 = T1;// [K]
+printf("Final temperature is %d K\n",T2);
+// From the ideal gas:
+// (P2*V2/T2) = (P1*V1/T1)
+// Since T2 = T1
+// P2*V2 = P1*V1
+P2 = P1*V1_by_V2;// [atm]
+printf("Final pressure is %.1f atm\n",P2);
+W = R*T1*log(1/V1_by_V2);// [L-atm]
+printf("Work done during the isothermal process is %.2f L-atm\n",W);
\ No newline at end of file diff --git a/647/CH2/EX2.13/Example2_13.sce b/647/CH2/EX2.13/Example2_13.sce new file mode 100755 index 000000000..9ff7a7e20 --- /dev/null +++ b/647/CH2/EX2.13/Example2_13.sce @@ -0,0 +1,56 @@ +clear;
+clc;
+
+// Example: 2.13
+// Page: 58
+
+printf("Example: 2.13 - Page: 58\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 5;// [kg]
+M = 29;// [kg/mol]
+T1 = 37 + 273;// [K]
+P1 = 101.33;// [kPa]
+T2 = 237 + 273;// [K]
+Cp = 29.1;// [J/mol K]
+Cv = 20.78;// [J/mol K]
+R = 8.314;// [J/K mol]
+//*****************//
+
+n = m/M;
+// From ideal gas equation:
+V1 = n*R*T1/P1;// [cubic m]
+
+// Isochoric Process:
+printf("Isochoric Process\n");
+// Volume = constant
+V2 = V1;// [cubic m]
+deltaU = n*Cv*(T2 - T1);// [kJ]
+// Since Volume is constant
+W = 0;
+Q = deltaU + W;// [kJ]
+// deltaH = deltaU + P*deltaV
+// deltaH = deltaU + n*R*deltaT
+deltaH = deltaU + n*R*(T2 - T1);// [kJ]
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Heat Supplied is %.2f kJ\n",Q);
+printf("Work done is %d kJ\n",W);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
+printf("\n");
+
+// Isobaric Process
+printf("Isobaric Process\n");
+// Since Pressure is constant.
+P2 = P1;// [kPa]
+deltaH = n*Cp*(T2 - T1);// [kJ]
+Qp = deltaH;// [kJ]
+// deltaU = deltaH - P*deltaV
+// From ideal gas equation:
+deltaU = deltaH - n*R*(T2 - T1);// [kJ]
+W = Qp - deltaU;// [kJ]
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Heat Supplied is %.2f kJ\n",Qp);
+printf("Work done is %.2f kJ\n",W);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
\ No newline at end of file diff --git a/647/CH2/EX2.14/Example2_14.sce b/647/CH2/EX2.14/Example2_14.sce new file mode 100755 index 000000000..e41d92d5c --- /dev/null +++ b/647/CH2/EX2.14/Example2_14.sce @@ -0,0 +1,84 @@ +clear;
+clc;
+
+// Example: 2.14
+// Page: 60
+
+printf("Example: 2.14 - Page: 60\n\n");
+
+// Solution
+
+//*****Data*****//
+n = 1;// [mol]
+T1 = 610;// [K]
+P1 = 10^6;// [N/square m]
+T2 = 310;// [K]
+P2 = 10^5;// [N/square m]
+Cv = 20.78;// [J/mol K]
+//*************//
+
+R = 8.314;// [J/K mol]
+
+// Step 1: Isothermal Expansion Of Ideal Gas:
+printf("Step 1: Isothermal Expansion Of Ideal Gas\n")
+T1 = 610;// [K]
+P1 = 10^6;// [N/square m]
+P2 = 10^5;// [N/square m]
+// Work done:
+W1 = 2.303*n*R*T1*log10(P1/P2);// [J/mol]
+// For isothermal expansion:
+delta_E1 = 0;// [J/mol]
+// From first law of thermodynamics:
+Q1 = delta_E1 + W1;// [J/mol]
+printf("delta_E for Step 1 is %d J/mol\n",delta_E1);
+printf("Q for step 1 is %.2f J/mol\n",Q1);
+printf("W for step 1 is %.2f J/mol\n",W1);
+printf("\n");
+
+// Step 2: Adiabatic Expansion of ideal gas:
+printf("Step 2: Adiabatic Expansion of ideal gas\n")
+Q2 = 0;// [J/mol]
+delta_E2 = Cv*(T2 - T1);// [J/mol]
+// From first law of thermodynamics:
+W2 = Q2 - delta_E2;// [J/mol]
+printf("delta_E for Step 2 is %d J/mol\n",delta_E2);
+printf("Q for step 2 is %.2f J/mol\n",Q2);
+printf("W for step 2 is %.2f J/mol\n",W2);
+printf("\n");
+
+// Step 3: Isothermal Compression Of Ideal Gas:
+printf("Step 3: Isothermal Compression Of Ideal Gas\n")
+T2 = 310;// [K]
+P1 = 10^5;// [N/square m]
+P2 = 10^6;// [N/square m]
+// Work done:
+W3 = 2.303*n*R*T2*log10(P1/P2);// [J/mol]
+// For isothermal expansion:
+delta_E3 = 0;// [J/mol]
+// From first law of thermodynamics:
+Q3 = delta_E3 + W3;// [J/mol]
+printf("delta_E for Step 3 is %d J/mol\n",delta_E3);
+printf("Q for step 3 is %.2f J/mol\n",Q3);
+printf("W for step 3 is %.2f J/mol\n",W3);
+printf("\n");
+
+// Step 4: Adiabatic Compression of ideal gas:
+printf("Step 4: Adiabatic Compression of ideal gas\n")
+T1 = 310;// [K]
+T2 = 610;// [K]
+Q4 = 0;// [J/mol]
+delta_E4 = Cv*(T2 - T1);// [J/mol]
+// From first law of thermodynamics:
+W4 = Q4 - delta_E4;// [J/mol]
+printf("delta_E for Step 4 is %d J/mol\n",delta_E4);
+printf("Q for step 4 is %.2f J/mol\n",Q4);
+printf("W for step 4 is %.2f J/mol\n",W4);
+printf("\n");
+
+// Net work done for the complete cycle:
+W = W1 + W2 + W3 + W4;// [J/mol]
+printf("Net Work done for the complete cycle is %.2f J/mol\n",W);
+
+// The efficiency of the cycle is given by:
+eta = 1- T1/T2;
+printf("The efficiency of the cycle is %.2f\n",eta);
\ No newline at end of file diff --git a/647/CH2/EX2.15/Example2_15.sce b/647/CH2/EX2.15/Example2_15.sce new file mode 100755 index 000000000..561a573ac --- /dev/null +++ b/647/CH2/EX2.15/Example2_15.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 2.15
+// Page: 61
+
+printf("Example: 2.15 - Page: 61\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 2.15 on page number 61 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 2.15 on page 61 of the book.")
\ No newline at end of file diff --git a/647/CH2/EX2.16/Example2_16.sce b/647/CH2/EX2.16/Example2_16.sce new file mode 100755 index 000000000..55e48cfd5 --- /dev/null +++ b/647/CH2/EX2.16/Example2_16.sce @@ -0,0 +1,66 @@ +clear;
+clc;
+
+// Example: 2.16
+// Page: 62
+
+printf("Example: 2.16 - Page: 62\n\n");
+
+// Solution
+
+//*****Data*****//
+P1 = 1;// [bar]
+T1 = 300;//[K]
+V1 = 24.92;// [cubic m/kmol]
+P2 = 10;// [bar]
+T2 = 300;// [K]
+Cp = 29.10;// [kJ/kmol K]
+Cv = 20.78;// [kJ/kmol K]
+R = 8.314;// [J/mol K]
+//**************//
+
+// Basis: 1 kmol of ideal gas:
+n = 1;
+V2 = P1*V1/P2;// [cubic m]
+
+// First Process:
+printf("First Process\n");
+// In the first step of the first process, the cooling of ga takes place at constant pressure.
+// Here the volume is reduced appreciably and consequently the temperature decreases.
+T_prime = T1*V2/V1;// [K]
+// Heat Requirement:
+Q1 = n*Cp*(T_prime - T1);// [kJ]
+deltaH1 = Q1;// [kJ]
+deltaU1 = deltaH1 - P1*(V2 - V1);// [kJ]
+// In the second step, the gas is heated at constant Volume:
+// V = constant
+Q2 = n*Cv*(T2 - T_prime);// [kJ]
+deltaU2 = Q2;// [kJ]
+deltaH2 = n*R*(T2 - T_prime);// [kJ]
+deltaU = deltaU1 + deltaU2;// [kJ]
+deltaH = deltaH1 + deltaH2;// [kJ]
+Q = Q1 + Q2;// [kJ]
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
+printf("Heat Requirement is %.2f kJ\n",Q);
+printf("\n");
+
+// Second Process:
+printf("Second Process\n");
+// In the first step of the second process, the gas is heated at constant volume.
+T_prime = T1*P2/P1;// [K]
+// Heat Requirement:
+Q1 = n*Cv*(T_prime - T1);// [kJ]
+deltaU1 = Q1;// [kJ]
+deltaH1 = n*R*(T_prime - T1);// [kJ]
+// In the second step, the gas is cooled at constant presure:
+// V = constant
+Q2 = n*Cp*(T2 - T_prime);// [kJ]
+deltaH2 = Q2;// [kJ]
+deltaU2 = deltaH2 - P1*(V2 - V1);// [kJ]
+deltaU = deltaU1 + deltaU2;// [kJ]
+deltaH = deltaH1 + deltaH2;// [kJ]
+Q = Q1 + Q2;// [kJ]
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
+printf("Heat Requirement is %.2f kJ\n",Q);
\ No newline at end of file diff --git a/647/CH2/EX2.17/Example2_17.sce b/647/CH2/EX2.17/Example2_17.sce new file mode 100755 index 000000000..21e84bbc0 --- /dev/null +++ b/647/CH2/EX2.17/Example2_17.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Example: 2.17
+// Page: 62
+
+printf("Example: 2.17 - Page: 64\n\n");
+
+// Solution
+
+//*****Data*****//
+D1 = 1;// [m]
+P1 = 120;// [kPa]
+P2 = 360;// [kPa]
+// P = k*D^3
+//***************//
+
+k = P1/D1^3;// [proportionality constant]
+D2 = (P2/k)^(1/3);// [m]
+// Work done by the gas inside the balloon can be estimated as:
+// W = integral(P*dV);
+// W = integral((k*D^3)*d((4/3)*%pi*r^3);
+// W = (%pi*k/6)*integral((D^3)*d(D^3));
+// W = (%pi*k/12)*(D2^6 - D1^6);
+W = (%pi*k/12)*(D2^6 - D1^6);// [kJ]
+printf("Workdone by the gas is %.2f kJ\n",W);
\ No newline at end of file diff --git a/647/CH2/EX2.18/Example2_18.sce b/647/CH2/EX2.18/Example2_18.sce new file mode 100755 index 000000000..38c14b6af --- /dev/null +++ b/647/CH2/EX2.18/Example2_18.sce @@ -0,0 +1,44 @@ +clear;
+clc;
+
+// Example: 2.18
+// Page: 65
+
+printf("Example: 2.18 - Page: 65\n\n");
+
+// Solution
+
+//*****Data*****//
+P1 = 10*100;// [kPa]
+T1 = 250;// [K]
+P2 = 1*100;// [kPa]
+T2 = 300;// [K]
+R = 8.314;// [J/mol K]
+Cv = 20.78;// [kJ/kmol K]
+Cp = 29.10;// [kJ/kmol K]
+//**********//
+
+V1 = R*T1/P1;// [cubic m]
+V2 = R*T2/P2;// [cubic m]
+
+// Calculation based on First Process:
+// In this constant-volume process, the initial pressure of 10 bar is reduced to a final pressure of 1 bar and consequently the temperature decreases.
+T_prime = P2*V1/R;// [K]
+deltaU1 = Cv*(T_prime - T1);// [kJ]
+deltaH1 = deltaU1 + V1*(P2 -P1);// [kJ]
+// Since V = constant
+W1 = 0;//[kJ]
+// By first law of thermodynamics:
+Q = W1 + deltaU1;// [kJ]
+
+// Calculation based on second process:
+// In this process, the gas is heated at constant pressure to the final temperature of T2.
+deltaH2 = Cp*(T2 - T_prime);// [kJ]
+deltaU2 = deltaH2 - P2*(V2 - V1);// [kJ]
+Q = deltaH2;// [kJ]
+W2 = Q - deltaU2;// [kJ]
+
+deltaU = deltaU1 + deltaU2;// [kJ]
+deltaH = deltaH1 + deltaH2;// [kJ]
+printf("Change in Inernal Enrgy is %.2f kJ\n",deltaU);
+printf("Change in Enthalpy is %.2f kJ\n",deltaH);
\ No newline at end of file diff --git a/647/CH2/EX2.19/Example2_19.sce b/647/CH2/EX2.19/Example2_19.sce new file mode 100755 index 000000000..188314831 --- /dev/null +++ b/647/CH2/EX2.19/Example2_19.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Example: 2.19
+// Page: 69
+
+printf("Example: 2.19 - Page: 69\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 273;// [K]
+T2 = 273 + 67;// [K]
+m_dot = 20000;// [kg/h]
+Ws = -1.5;// [hp]
+Q = -38000;// [kJ/min]
+Z = 20;// [m]
+Cp = 4.2;// [kJ/kg K]
+g = 9.81;// [m/second square]
+//***************//
+
+Q = Q*60/m_dot;// [kJ/kg]
+Ws = Ws*0.7457*3600/m_dot;// [kJ/kg]
+PE = g*Z*10^(-3);// [kJ/kg]
+// KE is assumed to be negligible.
+// For Steady Flow process: dE/dt = 0
+// From Eqn. 2.47:
+deltaH = Q - Ws - PE;// [kJ/kg]
+H1 = Cp*(T2 - T1);// [kJ/kg]
+H2 = H1 + deltaH;// [kJ/kg]
+
+// Now, the temperature of the tank can be determined as:
+T = (H2/Cp) + T1;// [K]
+printf("Tempertaure of water in the second tank is %.2f K\n",T);
\ No newline at end of file diff --git a/647/CH2/EX2.2/Example2_2.sce b/647/CH2/EX2.2/Example2_2.sce new file mode 100755 index 000000000..2304e0ae2 --- /dev/null +++ b/647/CH2/EX2.2/Example2_2.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 2.2
+// Page: 40
+
+printf("Example: 2.2 - Page: 40\n\n");
+
+// Solution
+
+//*****Data*****//
+U1 = 1000;// [kJ]
+Q = -600; // [kJ]
+W = -100;// [kJ]
+//************//
+
+// The system is considered to be a closed system. No mass transfer takes place across the system. The tank is rigid.
+// So, the kinetic and the potential energies is zero.
+// Therefore:
+// delta_E = delta_U + delta_PE + delta_KE
+// delta_E = delta_U
+// From the first law of thermodynamics:
+// Q = delta_U + W
+// delta_U = Q - W
+// U2 - U1 = Q - W
+U2 = U1 + Q - W;// [kJ]
+printf("The final internal energy of the fluid is %d kJ\n",U2);
\ No newline at end of file diff --git a/647/CH2/EX2.3/Example2_3.sce b/647/CH2/EX2.3/Example2_3.sce new file mode 100755 index 000000000..61495e31d --- /dev/null +++ b/647/CH2/EX2.3/Example2_3.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 2.3
+// Page: 40
+
+printf("Example: 2.3 - Page: 40\n\n");
+
+// Solution
+
+//*****Data*****//
+W = -3;// [hp]
+Q = -4000;// [kJ/h]
+//**************//
+
+// The work done by the stirrer on the system is given by
+W = W*745.7;// [W]
+// The amount of heat transferred to the suroundings can be expressed in terms of J/s:
+Q = Q*1000/3600;// [J/s]
+// From the first law of thermodynamics:
+// Q = delta_U - W
+delta_U = Q - W;// [J/s]
+printf("The change in the internal energy of the system would be %.2f J/s\n",delta_U);
\ No newline at end of file diff --git a/647/CH2/EX2.4/Example2_4.sce b/647/CH2/EX2.4/Example2_4.sce new file mode 100755 index 000000000..d72b423e0 --- /dev/null +++ b/647/CH2/EX2.4/Example2_4.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 2.4
+// Page: 41
+
+printf("Example: 2.4 - Page: 41\n\n");
+
+// Solution
+
+//*****Data*****//
+// From Fig. 2.4 (Page: 41)
+// For process A-1-B:
+Q1 = 60;// [kJ]
+W1 = 35;// [kJ]
+// For process A-2-B:
+W2 = 50;// [kJ]
+// For process B-3-A:
+W3 = -70;// [kJ]
+//************//
+
+// For process A-1-B:
+// The internal energy of the process A-1-B can be estimated as:
+// Q = delta_U + W
+delta_U = Q1 - W1;// [kJ]
+// For process A-2-B:
+Q2 = delta_U + W2;// [kJ]
+// For process B-3-A:
+Q3 = -delta_U + W3;// [kJ]
+printf("The amount of heat transferred from the system to the surroundings during process B-3-A is %d kJ\n",-Q3);
\ No newline at end of file diff --git a/647/CH2/EX2.5/Example2_5.sce b/647/CH2/EX2.5/Example2_5.sce new file mode 100755 index 000000000..727412cd3 --- /dev/null +++ b/647/CH2/EX2.5/Example2_5.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+// Example: 2.5
+// Page: 41
+
+printf("Example: 2.5 - Page: 41\n\n");
+
+// Solution
+
+//*****Data*****//
+// For constant pressure process 1-2:
+W12 = -100;// [kJ]
+Q12 = -50;// [kJ]
+// For constant volume process 2-3:
+Q23 = 80;// [kJ]
+// process 3-1: Adiabatic process
+//**************//
+
+// The internal energy of process 1-2 can be calculated as:
+delta_U12 = Q12 - W12;// [kJ]
+printf("Change in Internal Energy for process 1-2 is %d kJ\n",delta_U12);
+// For the process 2-3:
+// As the process is constant volume process:
+W23 = 0;// [kJ]
+delta_U23 = Q23 - W23;// [kJ]
+printf("Change in Internal Energy for process 2-3 is %d kJ\n",delta_U23);
+
+// For process 3-1:
+// Since the process is adiabatic, ther is no heat transfer between the system and the surrounding.
+Q31 = 0;// [kJ]
+// For a cyclic process, the internal energy change is zero.
+// delta_U12 + delta_U23 + delta_U31 = 0
+delta_U31 = -(delta_U12 + delta_U23);// [kJ]
+// Putting the value of delta_U31:
+W31 = Q31 - delta_U31;// [kJ]
+printf("Change in Internal Energy for process 3-1 is %d kJ\n",delta_U31);
+printf("The work done during the adiabatic process is %d kJ\n",W31);
\ No newline at end of file diff --git a/647/CH2/EX2.6/Example2_6.sce b/647/CH2/EX2.6/Example2_6.sce new file mode 100755 index 000000000..2b39429ad --- /dev/null +++ b/647/CH2/EX2.6/Example2_6.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Example: 2.6
+// Page: 44
+
+printf("Example: 2.6 - Page: 44\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 1;// [kg]
+Temp = 373;// [K]
+P = 101325;// [N/square m]
+V_Liquid = 1.04*10^(-3);// [cubic m/kg]
+V_Vapour = 1.673;// [cubic m/kg]
+Q = 2257;// [kJ]
+//**************//
+
+// Work done due to expansion:
+Wexpansion = P*(V_Vapour - V_Liquid);// [N-m]
+deltaU = Q - Wexpansion/1000;// [kJ]
+deltaH = deltaU + Wexpansion/1000;// [kJ]
+
+printf("Change in Internal Energy is %.2f kJ\n",deltaU);
+printf("Change in enthalpy is %d kJ\n",deltaH);
\ No newline at end of file diff --git a/647/CH2/EX2.7/Example2_7.sce b/647/CH2/EX2.7/Example2_7.sce new file mode 100755 index 000000000..003950ff8 --- /dev/null +++ b/647/CH2/EX2.7/Example2_7.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 2.7
+// Page: 45
+
+printf("Example: 2.7 - Page: 45\n\n");
+
+// Solution
+
+//*****Data*****//
+n = 1;// [mol]
+Temp = 353;// [K]
+P = 1;// [atm]
+Hv = 380;// [J/g]
+Mwt = 78;// [g/mol]
+R = 8.314;// [J/K mol]
+//*************//
+
+Q = Hv*Mwt;// [J/mol]
+// Since Vv >> Vl:
+// P*(Vv - Vl) = P*Vv =n*R*Temp
+Wexpansion = n*R*Temp;// [J]
+// By first law of thermodynamics:
+deltaU = Q - Wexpansion;// [J]
+deltaH = deltaU + Wexpansion;// [J]
+
+printf("Change in Internal Energy is %.2f J\n",deltaU);
+printf("Change in Enthalpy is %d J\n",deltaH);
+printf("Amount of Heat supplied is %d J\n",Q);
+printf("Work done is %.2f J\n",Wexpansion);
\ No newline at end of file diff --git a/647/CH2/EX2.8/Example2_8.sce b/647/CH2/EX2.8/Example2_8.sce new file mode 100755 index 000000000..ba049a1df --- /dev/null +++ b/647/CH2/EX2.8/Example2_8.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 2.8
+// Page: 45
+
+printf("Example: 2.8 - Page: 45\n\n");
+
+// Solution
+
+//*****Data*****//
+deltaU = 200;// [cal]
+Vinit = 10;// [L]
+Vfinal = 50;// [L]
+deff('[P] = f(V)','P = 10/V');
+//**************//
+
+// By definition of enthalpy:
+// deltaQ = deltaU + PdV
+deltaQ = deltaU + integrate('f(V)','V',Vinit,Vfinal)*24.2;// [cal]
+printf("Change in enthalpy is %f cal\n",deltaQ);
\ No newline at end of file diff --git a/647/CH2/EX2.9/Example2_9.sce b/647/CH2/EX2.9/Example2_9.sce new file mode 100755 index 000000000..83d22b566 --- /dev/null +++ b/647/CH2/EX2.9/Example2_9.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Example: 2.9
+// Page: 48
+
+printf("Example: 2.9 - Page: 48\n\n");
+
+// Solution
+
+//*****Data*****//
+m_water = 1;// [kg]
+Cv = 4.18;// [kJ/kg K]
+m_stirrer = 40;// [kg]
+h = 25;// [m]
+g = 9.81;// [m/square s]
+//***************//
+
+// Since the system is thermally insulated:
+// Q = 0
+// From the first law of thermodynamics:
+// dQ = dE + dW
+// As E = U + Ek +Ep and Ek = Ep = 0
+// dQ = dU + dW
+dT = g*h/Cv;// [K]
+printf("Rise in Temperature is %.2f K\n",dT);
\ No newline at end of file diff --git a/647/CH3/EX3.1/Example3_1.sce b/647/CH3/EX3.1/Example3_1.sce new file mode 100755 index 000000000..4b7c21a9a --- /dev/null +++ b/647/CH3/EX3.1/Example3_1.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 3.1
+// Page: 88
+
+printf("Example: 3.1 - Page: 88\n\n");
+
+// Solution
+
+//*****Data*****//
+P = 2*10^5;// [Pa]
+T = 273 + 37;// [K]
+R = 8.314;// [J/mol K]
+//****************//
+
+// Since the gas behaves ideally:
+V = R*T/P;// [cubic metres]
+printf("Molar Volume of the gas is %.2e cubic metres",V);
\ No newline at end of file diff --git a/647/CH3/EX3.10/Example3_10.sce b/647/CH3/EX3.10/Example3_10.sce new file mode 100755 index 000000000..7f8a27f88 --- /dev/null +++ b/647/CH3/EX3.10/Example3_10.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+// Example: 3.10
+// Page: 103
+
+printf("Example: 3.10 - Page: 103\n\n");
+
+// Solution
+
+//*****Data*****//
+beeta = 1.487*10^(-3);// [1/OC]
+alpha = 62*10^(-6);// [1/bar]
+V1 = 1.287;// [cubic cm /g]
+//************//
+
+// Solution (a)
+// The value of derivative (dP/dT) at constant V:
+// dV/V = beeta*dT - alpha*dP
+// dV = 0
+// dP/dT = beeta/alpha
+// Value = dP/dT
+Value = beeta/alpha;// [bar/OC]
+printf("Value of derivative is %.2f bar/OC\n",Value);
+
+// Solution (b)
+P1 = 1;// [bar]
+T1 = 20;// [OC]
+T2 = 30;// [OC]
+// Applying the same equation:
+P2 = P1 +(beeta/alpha)*(T2 - T1);// [bar]
+printf("The pressure generated by heating at constant Volume is %.2f Pa\n",P2);
+
+// Solution (c)
+T2 = 0;// [OC]
+T1 = 20;// [OC]
+P2 = 10;// [bar]
+P1 = 1;// [bar]
+// The change in Volume can be obtained as:
+V2 = V1*exp((beeta*(T2 - T1)) - alpha*(P2 - P1));// [cubic cm/g]
+deltaV = V2 - V1;// [cubic cm/g]
+printf("The change in Volume is %.3f cubic cm/g\n",deltaV);
\ No newline at end of file diff --git a/647/CH3/EX3.11/Example3_11.sce b/647/CH3/EX3.11/Example3_11.sce new file mode 100755 index 000000000..6ac349201 --- /dev/null +++ b/647/CH3/EX3.11/Example3_11.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 3.11
+// Page: 107
+
+printf("Example: 3.11 - Page: 107\n\n");
+
+// Solution
+
+//*****Data*****//
+Tc = 513.9;// [K]
+Pc = 61.48*10^5;// [Pa]
+//************//
+
+Tr = 0.7;
+T = Tr*Tc - 273.15;// [OC]
+P_sat = 10^(8.112 - (1592.864/(T + 226.184)));// [mm Hg]
+P_sat = P_sat*101325/760;// [Pa]
+Pr_sat = P_sat/Pc;// [Pa]
+omega = -1 - log10(Pr_sat);// [Acentric factor]
+printf("Acentric factor is %.4f",omega);
\ No newline at end of file diff --git a/647/CH3/EX3.2/Example3_2.sce b/647/CH3/EX3.2/Example3_2.sce new file mode 100755 index 000000000..80b412c52 --- /dev/null +++ b/647/CH3/EX3.2/Example3_2.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 3.2
+// Page: 89
+
+printf("Example: 3.2 - Page: 89\n\n");
+
+// Solution
+
+//*****Data*****//
+V1 = 8;// [cubic m]
+P1 = 300;// [kPa]
+V2 = 2;// [cubic m]
+//**************//
+
+// Apptying the ideal gas Eqn. & since the Temperature remains constant:
+P2 = P1*V1/V2;// [kPa]
+printf("The pressure of air after compression is %d kPa\n",P2);
\ No newline at end of file diff --git a/647/CH3/EX3.3/Example3_3.sce b/647/CH3/EX3.3/Example3_3.sce new file mode 100755 index 000000000..f7d8a16ca --- /dev/null +++ b/647/CH3/EX3.3/Example3_3.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 3.3
+// Page: 89
+
+printf("Example: 3.3 - Page: 89\n\n");
+
+// Solution
+
+//*****Data*****//
+V1 = 6;// [cubic m]
+P1 = 500;// [kPa]
+R = 0.287;// [kJ/kg K]
+//*************//
+
+// Applying the charectarstic equation to the gas initially:
+// P1*V1 = m1*R*T1.......................................(i)
+// Applying the charectarstic equation to the gas which was left in the vessel after one-fifth of the gas has been removed:
+// P2*V2 = m2*R*T2.......................................(ii)
+// V2 = V1;
+// T2 = T1;
+// m2 = (4/5)*m1;
+// Eqn (ii) becomes:
+// P2*V1 = (4/5)*m1*R*T1..................................(iii)
+// Dividing eqn (i) by eqn (iii), we get:
+P2 = (4/5)*P1;// [kPa]
+printf("The pressure of the remaining air is %d kPa\n",P2);
\ No newline at end of file diff --git a/647/CH3/EX3.4/Example3_4.sce b/647/CH3/EX3.4/Example3_4.sce new file mode 100755 index 000000000..b2dff7a14 --- /dev/null +++ b/647/CH3/EX3.4/Example3_4.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// Example: 3.4
+// Page: 90
+
+printf("Example: 3.4 - Page: 90\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 450 + 273;// [K]
+P1 = 3;// [bar]
+//***************//
+
+// Soluton(a)
+// From Fig. 3.7, (Page 90)
+// Since the weight remains the same, therefore, the final pressure is equal to the initial pressure.
+// Therefore it is a constant pressure process.
+P2 = P1;// [bar]
+// Volumetric Ratio:
+V2_by_V1 = 2.5/(2.5 + 2.5);
+// Applying ideal gas law & P1 = P2
+T2 = T1*V2_by_V1;// [K]
+printf("Final Temperature of the air when the piston reaches stop is %.1f K\n",T2);
+
+// Solution (b)
+// When the piston rests ot the stops, the pressure exerted by the weight, air & the atmosphere will be different. But there will beno further decrease in volume.
+// This is a constant volume process.
+T3 = 273 + 30;// [K]
+// Applying ideal gas law & V2 = V3
+P3 = T3*P2/T2;// [bar]
+printf("Pressure of air inside the cylinder is %.2f bar\n",P3);
\ No newline at end of file diff --git a/647/CH3/EX3.5/Example3_5.sce b/647/CH3/EX3.5/Example3_5.sce new file mode 100755 index 000000000..cf403fe21 --- /dev/null +++ b/647/CH3/EX3.5/Example3_5.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 3.5
+// Page: 95
+
+printf("Example: 3.5 - Page: 95\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 1.373;// [kg]
+P = 1.95*10^(6);// [Pa]
+V = 0.1;// [cubic m]
+a = 422.546*10^(-3);// [cubic m/square mol]
+b = 37*10^(-6);// [cubic m/mol]
+M = 17*10^(-3);// [kg/mol]
+R = 8.314;// [J/mol K]
+//****************//
+
+n = m/M;// [moles]
+Vm = V/n;// [molar volume, cubic m]
+// Applying Van der Waals equation of state:
+T = (P + (a/Vm^2))*((Vm - b)/R);// [K]
+printf("The temperature at which ammonia exists in the cylinder is %.1f K\n",T)
\ No newline at end of file diff --git a/647/CH3/EX3.6/Example3_6.sce b/647/CH3/EX3.6/Example3_6.sce new file mode 100755 index 000000000..45ac1647d --- /dev/null +++ b/647/CH3/EX3.6/Example3_6.sce @@ -0,0 +1,52 @@ +clear;
+clc;
+
+// Example: 3.6
+// Page: 96
+
+printf("Example: 3.6 - Page: 96\n\n");
+
+// Solution
+
+//*****Data*****//
+P = 15*10^5;// [Pa]
+T = 773;// [K]
+R = 8.314;// [J/mol K]
+//**************//
+
+// Solution (a)
+printf("Ideal Equation of State\n")
+// Applying ideal Eqn. of State:
+Vm = R*T/P;// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (b)
+printf("Van der Wall Equation of State\n");
+a = 0.2303;// [Nm^4/square mol]
+b = 4.3073*10^(-5);// [cubic m/mol]
+deff('[y] = f1(Vm)','y = P - (R*T/(Vm-b)) + (a/Vm^2)');
+Vm = fsolve(Vm,f1);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+//Solution (c)
+printf("Virial Equation of State\n");
+// Z = 1 + B/V
+// (P*V/(R*T)) = (1 + B/V)
+// V^2 - V*R*T/P - B*R*T/P = 0
+B = 1.3697*10^(-5);// [cubic m/mol]
+deff('[y] = f2(Vm)','y = Vm^2 - (Vm*R*T/P) - (B*R*T/P)');
+Vm = fsolve(7,f2);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (d)
+printf("Redlich Kwong Equation of State\n");
+Tc = 190.6;// [K]
+Pc = 45.99*10^5;// [Pa]
+a = 0.4278*R^2*Tc^2.5/Pc;// [N/m^4 square mol]
+b = 0.0867*R*Tc/Pc;// [cubic m/mol]
+deff('[y] = f3(Vm)','y = P - (R*T/(Vm - b)) + (a/((T^0.5)*Vm*(Vm+b)))');
+Vm = fsolve(Vm,f3);// [cubic m/mol]
+printf("Molar Volume of the gas is %.3e cubic m/mol\n",Vm);
\ No newline at end of file diff --git a/647/CH3/EX3.7/Example3_7.sce b/647/CH3/EX3.7/Example3_7.sce new file mode 100755 index 000000000..a1e824d8e --- /dev/null +++ b/647/CH3/EX3.7/Example3_7.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 3.7
+// Page: 101
+
+printf("Example: 3.7 - Page: 101\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 3.7 on page number 101 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 3.7 on page 101 of the book.")
\ No newline at end of file diff --git a/647/CH3/EX3.8/Example3_8.sce b/647/CH3/EX3.8/Example3_8.sce new file mode 100755 index 000000000..220e5c9f5 --- /dev/null +++ b/647/CH3/EX3.8/Example3_8.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 3.8
+// Page: 101
+
+printf("Example: 3.8 - Page: 101\n\n");
+
+// Solution
+
+//*****Data*****//
+T = 500;// [K]
+P = 8*10^6;// [Pa]
+R = 8.314;// [J/mol K]
+//*************//
+
+// Solution (a)
+// By ideal gas equation of state:
+printf("Ideal Equation of State\n")
+Vm = R*T/P;// [cubic m/mol]
+printf("Molar Volume of gas is %.3e cubic m/mol\n",Vm);
+printf("\n");
+
+// Solution (b)
+// By Virial Equation of State:
+printf("Virial Equation of State\n");
+B = -0.265*10^(-3);// [cubic m/mol]
+C = 0.3025*10^(-7);// [m^6/square mol]
+deff('[y] = f(Vm)','y = (P*Vm/(R*T)) - 1 -(B/Vm) - (C/Vm^2)');
+Vm = fsolve(Vm,f);
+printf("Molar Volume of gas is %.2e cubic m/mol\n",Vm);
\ No newline at end of file diff --git a/647/CH3/EX3.9/Example3_9.sce b/647/CH3/EX3.9/Example3_9.sce new file mode 100755 index 000000000..8549376d6 --- /dev/null +++ b/647/CH3/EX3.9/Example3_9.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 3.9
+// Page: 102
+
+printf("Example: 3.9 - Page: 102\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 3.9 on page number 102 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 3.9 on page 102 of the book.")
\ No newline at end of file diff --git a/647/CH4/EX4.1/Example4_1.sce b/647/CH4/EX4.1/Example4_1.sce new file mode 100755 index 000000000..9329c044f --- /dev/null +++ b/647/CH4/EX4.1/Example4_1.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 4.1
+// Page: 118
+
+printf("Example: 4.1 - Page: 118\n\n");
+
+// Solution
+
+//*****Data*****//
+Qp = -327;// [kcal]
+T = 27 + 273;// [K]
+R = 2*10^(-3);// [kcal/K mol]
+//*************//
+
+// The reaction involved is:
+// C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l)
+deltan = 2 - 3;
+Qv = Qp - deltan*R*T;// [kcal]
+printf("Value of Qv is %.2f kcal\n",Qv);
\ No newline at end of file diff --git a/647/CH4/EX4.10/Example4_10.sce b/647/CH4/EX4.10/Example4_10.sce new file mode 100755 index 000000000..d8d3ed8f5 --- /dev/null +++ b/647/CH4/EX4.10/Example4_10.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 4.10
+// Page: 129
+
+printf("Example: 4.10 - Page: 129\n\n");
+
+// Solution
+
+//*****Data*****//
+T2 = 1273;// [K]
+T1 = 300;// [K]
+deltaH_300 = -11030;// [cal/mol]
+//*************//
+
+// The chemical reaction involved is:
+// N2 + 3H2 = 2NH3
+// (1/2)N2 + (3/2)H2 = NH3
+// deltaH_1273 = deltaH_300 + integrate('Cp_NH3(T) - (1/2)*Cp_N2(T) - (1/2)*Cp_H2(T)','T',1273,300);
+deltaH_1273 = deltaH_300 + integrate('(6.2 + 7.8*10^(-3)*T - 7.2*10^(-6)*T^2) - (1/2)*(6.45 + 1.4*10^(-3)*T) - (1/2)*(6.94 - 0.2*10^(-3)*T)','T',1273,300);// [cal]
+printf("Heat of formation at 1273 K is %d cal",deltaH_1273);
\ No newline at end of file diff --git a/647/CH4/EX4.11/Example4_11.sce b/647/CH4/EX4.11/Example4_11.sce new file mode 100755 index 000000000..820cdc6c3 --- /dev/null +++ b/647/CH4/EX4.11/Example4_11.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 4.11
+// Page: 130
+
+printf("Example: 4.11 - Page: 130\n\n");
+
+// Solution
+
+//*****Data*****//
+CO2 = 13.4;// [percent by volume]
+N2 = 80.5;// [percent by volume]
+O2 = 6.1;// [percent by volume]
+//*************//
+
+// Basis : 100 cubic m of flue gas.
+Vol_N2_flue = N2;// [Volume of Nitrogen in flue gas, cubic m]
+Vol_O2_flue = O2;// [Volume of O2 in flue gas, cubic m]
+Vol_Air = N2/0.79;// [Volume of air supplied, cubic m]
+Vol_O2 = Vol_Air*0.21;// [Volume of O2 in air supply, cubic m]
+Vol_O2_cumbustion = Vol_O2 - Vol_O2_flue;// [Volume of O2 used up in cumbustion of the fuel, cubic m]
+Excess_Air = Vol_O2_flue/Vol_O2_cumbustion * 100;// [percent of excess air supplied]
+printf("Percent of excess air supplied is %.1f %%",Excess_Air);
\ No newline at end of file diff --git a/647/CH4/EX4.2/Example4_2.sce b/647/CH4/EX4.2/Example4_2.sce new file mode 100755 index 000000000..f757a2a6f --- /dev/null +++ b/647/CH4/EX4.2/Example4_2.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 4.2
+// Page: 119
+
+printf("Example: 4.2 - Page: 119\n\n");
+
+// Solution
+
+//*****Data*****//
+// Mg + (1/2)O2 = MgO ...............(1)
+deltaH1 = -610.01;// [kcal]
+// 2Fe + (3/2)O2 = Fe2O3 ............(2)
+deltaH2 = -810.14;// [kcal]
+//*************//
+
+// 3Mg + Fe2O3 = 3MgO + 2Fe .........(3)
+// Multiplying (1) by 3 and substracting from (2), we get (3):
+deltaH = 3*deltaH1 - deltaH2;// [kcal]
+printf("Heat produced in the reaction is %.1f kcal\n",deltaH);
\ No newline at end of file diff --git a/647/CH4/EX4.3/Example4_3.sce b/647/CH4/EX4.3/Example4_3.sce new file mode 100755 index 000000000..1809a6117 --- /dev/null +++ b/647/CH4/EX4.3/Example4_3.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 4.3
+// Page: 121
+
+printf("Example: 4.3 - Page: 121\n\n");
+
+// Solution
+
+//*****Data*****//
+// 2H2(g) + O2(g) ---------------> 2H2O .....................(1)
+deltaH1 = -241.8*2;// [kJ/gmol H2]
+// C(graphite) + O2(g) =---------> CO2(g) ...................(2)
+deltaH2 = -393.51;// [kJ/gmol C]
+// CH4(g) + 2O2(g) ---------------> CO2(g) + 2H2O(l) ........(3)
+deltaH3 = -802.36;// [kJ/mol CH4]
+//*************//
+
+// For standard heat of formation of methane, (a) + (b) - (c)
+// C + 2H2 ------------------------> CH4
+deltaHf = deltaH1 + deltaH2 - deltaH3;// [kJ/gmol]
+printf("The standard heat of formation of methane is %.2f kJ/gmol\n",deltaHf);
\ No newline at end of file diff --git a/647/CH4/EX4.4/Example4_4.sce b/647/CH4/EX4.4/Example4_4.sce new file mode 100755 index 000000000..088d98ae9 --- /dev/null +++ b/647/CH4/EX4.4/Example4_4.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 4.4
+// Page: 122
+
+printf("Example: 4.4 - Page: 122\n\n");
+
+// Solution
+
+//*****Data*****//
+deltaH_C6H12O6 = -1273;// [kcal]
+deltaH_C2H5OH = -277.6;// [kcal]
+deltaH_CO2 = -393.5;// [kcal]
+deltaH_H2O = -285.8;// [kcal]
+//**************//
+
+// C6H12O6(s) = 2C2H5OH(l) + 2CO2(g) ..........................(A)
+deltaH_A = 2*deltaH_C2H5OH + 2*deltaH_CO2 - deltaH_C6H12O6;// [kJ]
+// C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l) ...................(B)
+deltaH_B = 6*deltaH_CO2 + 6*deltaH_H2O - deltaH_C6H12O6;// [kJ]
+printf("Energy supplied by reaction A is %.1f kJ\n",deltaH_A);
+printf("Energy supplied by reaction B is %.1f kJ\n",deltaH_B);
+if deltaH_A < deltaH_B
+ printf("Reaction A supplies more energy to the organism\n");
+else
+ printf("Reaction B supplies more energy to the organism\n");
+end
\ No newline at end of file diff --git a/647/CH4/EX4.5/Example4_5.sce b/647/CH4/EX4.5/Example4_5.sce new file mode 100755 index 000000000..ad2f618be --- /dev/null +++ b/647/CH4/EX4.5/Example4_5.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 4.5
+// Page: 122
+
+printf("Example: 4.5 - Page: 122\n\n");
+
+// Solution
+
+//*****Data*****//
+// Zn + S = ZnS ....................................................(A)
+deltaH_A = -44;// [kcal/kmol]
+// ZnS + 3O2 = 2ZnO + 2SO2 .........................................(B)
+deltaH_B = -221.88;// [kcal/kmol]
+// 2SO2 + O2 = 2SO3 ................................................(C)
+deltaH_C = -46.88;// [kcal/kmol]
+// ZnO + SO3 = ZnSO4 ...............................................(D)
+deltaH_D = -55.10;// [kcal/kmol]
+//***************//
+
+// Multiplying (A) by 2 & (D) by (2) and adding (A), (B), (C) & (D)
+// Zn + S + 2O2 = ZnSO4
+deltaH = 2*deltaH_A + deltaH_B + deltaH_C + 2*deltaH_D;// [kcal/kmol for 2 kmol of ZnSO4]
+printf("Heat of formation of ZnSO4 is %.2f kcal/kmol\n",deltaH/2);
\ No newline at end of file diff --git a/647/CH4/EX4.6/Example4_6.sce b/647/CH4/EX4.6/Example4_6.sce new file mode 100755 index 000000000..11dbbb0ea --- /dev/null +++ b/647/CH4/EX4.6/Example4_6.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 4.6
+// Page: 124
+
+printf("Example: 4.6 - Page: 124\n\n");
+
+// Solution
+
+//*****Data*****//
+// HC : Heat of Combustion
+HC_NH3 = -90.6;// [kcal]
+HC_H2 = -68.3;// [kcal]
+//*************//
+
+// Heat of combustion of NH3:
+// 2NH3 + 3O = N2 + 3H2O ............................ (A)
+// Heat of combustion of H2:
+// H2 + O = H2O ..................................... (B)
+// Multiplying (B) by 3 & substracting from (A), we get:
+// 2NH3 = N2 + 3H2 .................................. (C)
+// Hf : Heat of Formation
+Hf_NH3 = -(2*HC_NH3 - 3*HC_H2)/2;// [kcal]
+printf("Standard Heat of formation of NH3 is %.1f kcal",Hf_NH3);
\ No newline at end of file diff --git a/647/CH4/EX4.7/Example4_7.sce b/647/CH4/EX4.7/Example4_7.sce new file mode 100755 index 000000000..fbfbde81c --- /dev/null +++ b/647/CH4/EX4.7/Example4_7.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Example: 4.7
+// Page: 125
+
+printf("Example: 4.7 - Page: 125\n\n");
+
+// Solution
+
+//*****Data*****//
+// HC : Heat of Combustion
+HC_C2H2 = -310600; // [cal]
+//**************//
+
+// C2H2 + (5/2)O2 = 2CO2 + H2O
+Q = -HC_C2H2;// [cal]
+// The gases present in the flame zone after combustion are carbon dioxide, water vapor and the unreacted nitrogen of the air.
+// Since (5/2) mole of oxygen were required for combustion, nitrogen required would be 10 mol.
+// Hence the composition of the resultant gas would be 2 mol CO2, 1 ol H2 & 10 mol N2.
+// Q = integrate('Cp(T)','T',T,298);
+// On integrating we get:
+// Q = 84.52*(T - 298) + 18.3*10^(-3)*(T^2 - 298^2)
+deff('[y] = f(T)','y = Q - 84.52*(T - 298) - 18.3*10^(-3)*(T^2 - 298^2)');
+T = fsolve(7,f);// [K]
+printf("The maximum attainable temperature is %.1f K",T);
\ No newline at end of file diff --git a/647/CH4/EX4.8/Example4_8.sce b/647/CH4/EX4.8/Example4_8.sce new file mode 100755 index 000000000..1c8a60d43 --- /dev/null +++ b/647/CH4/EX4.8/Example4_8.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Example: 4.8
+// Page: 126
+
+printf("Example: 4.8 - Page: 126\n\n");
+
+// Solution
+
+//*****Data*****//
+Cp_CO2 = 54.56;// [kJ/mol K]
+Cp_O2 = 35.20;// [kJ/mol K]
+Cp_steam = 43.38;// [kJ/mol K]
+Cp_N2 = 33.32;// [kJ/mol K]
+// 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(g)
+deltaH_273 = -1560000;// [kJ/kmol]
+//************//
+
+// Since the air is 25% in excess of the amount required,the combustion may be written as:
+// C2H6(g) + (7/2)O2(g) = 2CO2(g) + 3H2O(g)
+// 25% excess air is supplied.
+// Since the air contains N2 = 79% and O2 = 21%
+// C2H6(g) + 3.5O2(g) + 0.25*3.5O2(g) + (4.375*(79/21))N2 = 2CO2 + 3H2O + 0.875O2 + 16.46N2 .................. (A)
+// Considering the reaction (A),
+// Amount of O2:
+O2 = 3.5 + 3.5*0.25;// [mol]
+// Amount of N2 required:
+N2 = 4.375*(79/21);// [mol]
+// Let the initial temperature of ethane and air be 0 OC and the temperature of products of combustion be T OC
+// Since heat librated by combustion = heat accumulated by combustion products
+Q = -deltaH_273;// [kJ/mol K]
+T = Q/(2*Cp_CO2 + 3*Cp_steam + 0.875*Cp_O2 + N2*Cp_N2);// [OC]
+printf("The theoretical temperature of combustion is %d degree Celsius",T);
\ No newline at end of file diff --git a/647/CH4/EX4.9/Example4_9.sce b/647/CH4/EX4.9/Example4_9.sce new file mode 100755 index 000000000..c6c94db83 --- /dev/null +++ b/647/CH4/EX4.9/Example4_9.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 4.9
+// Page: 129
+
+printf("Example: 4.9 - Page: 129\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 273;// [K]
+T2 = 253;// [K]
+deltaH_273 = 1440;// [cal/mol]
+Cp = 8.7;// [cal/mol]
+//**************//
+
+deltaH_253 = deltaH_273 + Cp*(T2 - T1);// [cal/mol]
+printf("Laten heat of ice at -20 OC is %d cal/mol\n",deltaH_253);
\ No newline at end of file diff --git a/647/CH5/EX5.1/Example5_1.sce b/647/CH5/EX5.1/Example5_1.sce new file mode 100755 index 000000000..4dffa7e93 --- /dev/null +++ b/647/CH5/EX5.1/Example5_1.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 5.1
+// Page: 150
+
+printf("Example: 5.1 - Page: 150\n\n");
+
+// Solution
+
+//*****Data*****//
+Th = 550 + 273;// [K]
+Tl = 27 + 273;// [K]
+//************//
+
+// The theoretical efficiency of a heat engine is given by:
+// eta = Net Work Output/Net Work Input
+// eta = Wnet/Qin
+// eta = (Qin - Qout)/Qin = (Th - Tl)/Th
+eta = (Th - Tl)/Th;
+printf("The theoretical efficiency of heat engine is %.1f %%",eta * 100)
\ No newline at end of file diff --git a/647/CH5/EX5.10/Example5_10.sce b/647/CH5/EX5.10/Example5_10.sce new file mode 100755 index 000000000..35e98bd87 --- /dev/null +++ b/647/CH5/EX5.10/Example5_10.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 5.10
+// Page: 164
+
+printf("Example: 5.10 - Page: 164\n\n");
+
+// Solution
+
+//*****Data*****//
+n = 5;// [moles]
+T1 = 300;// [K]
+T2 = 400;// [K]
+P1 = 3;// [bars]
+P2 = 12;// [bars]
+Cp = 26.73;// [J/mol K]
+R = 8.314;// [K/mol K]
+//*************//
+
+deltaS = n*((Cp*log(T2/T1)) + (R*log(P1/P2)));// [kJ/K]
+printf("Change in Entropy is %f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.11/Example5_11.sce b/647/CH5/EX5.11/Example5_11.sce new file mode 100755 index 000000000..9406ceceb --- /dev/null +++ b/647/CH5/EX5.11/Example5_11.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 5.11
+// Page: 166
+
+printf("Example: 5.11 - Page: 166\n\n");
+
+// Solution
+
+//*****Data*****//
+N = 1;// [kmol]
+xA = 0.21;// [for Oxygen]
+xB = 0.79;// [for Nitrogen]
+R = 8.314;// [kJ/kmol K]
+//*************//
+
+deltaS = - (N*R*(xA*log(xA) + xB*log(xB)));// [kJ/mol K]
+printf("Entropy Change is %.2f kJ/kmol K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.12/Example5_12.sce b/647/CH5/EX5.12/Example5_12.sce new file mode 100755 index 000000000..7c49e3740 --- /dev/null +++ b/647/CH5/EX5.12/Example5_12.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 5.12
+// Page: 167
+
+printf("Example: 5.12 - Page: 167\n\n");
+
+// Solution
+
+//*****Data*****//
+Vol_O2 = 5.6;// [L]
+Vol_H2 = 16.8;// [L]
+R = 1.987;// [cal/mol K]
+//*************//
+
+xA = Vol_O2/22.4;// [mole fraction O2]
+xB = Vol_H2/22.4;// [mle fraaction H2]
+N = xA + xB;// [total number of moles]
+// From Eqn. 5.21:
+deltaS = - (N*R*(xA*log(xA) + xB*log(xB)));// [cal/K]
+printf("Change in Entropy is %.3f cal/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.13/Example5_13.sce b/647/CH5/EX5.13/Example5_13.sce new file mode 100755 index 000000000..e0241e0be --- /dev/null +++ b/647/CH5/EX5.13/Example5_13.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 5.13
+// Page: 168
+
+printf("Example: 5.13 - Page: 168\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 80;// [mass of Argon, g]
+T1 = 300;// [K]
+T2 = 500;// [K]
+Cv = 0.3122;// [kJ/kg K]
+//**************//
+
+Mw = 40;// [Molecular Weight of Argon]
+n = m/Mw;// [moles]
+deltaS = n*Cv*log(T2/T1);// [kJ/K]
+printf("Entropy Change is %.3f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.14/Example5_14.sce b/647/CH5/EX5.14/Example5_14.sce new file mode 100755 index 000000000..4f58e1243 --- /dev/null +++ b/647/CH5/EX5.14/Example5_14.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 5.14
+// Page: 168
+
+printf("Example: 5.14 - Page: 168\n\n");
+
+// Solution
+
+//*****Data*****//
+deltaS = 1;// [kJ/kg K]
+Cv = 0.918;// [kJ/kg K]
+T1 = 273 + 18;// [K]
+//*************//
+
+// Let the upper temperature be T.
+// deltaS = integrate('Cv/T','T',T1,T);
+// deltaS = Cv*log(T/T1)
+T = T1*exp(deltaS/Cv);// [K]
+printf("The upper temperature of the process is %.3f K",T);
\ No newline at end of file diff --git a/647/CH5/EX5.15/Example5_15.sce b/647/CH5/EX5.15/Example5_15.sce new file mode 100755 index 000000000..b0666eaa2 --- /dev/null +++ b/647/CH5/EX5.15/Example5_15.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 5.15
+// Page: 169
+
+printf("Example: 5.15 - Page: 169\n\n");
+
+// Solution
+
+//*****Data*****//
+m1 = 5;// [kg]
+m2 = 20;// [kg]
+C = 4.2;// [kJ/kg K]
+T1 = 350;// [K]
+T2 = 250;// [K]
+//**************//
+
+// Suppose the final temperature is T
+deff('[y] = f(T)','y = m1*C*(T1 - T) - m2*C*(T - T2)');
+T = fsolve(7,f);// [K]
+// Change in entropy of Hot Water:
+deltaS1 = m1*C*integrate('(1/T)','T',T1,T);// [kJ/K]
+// Change in Entopy of Hot Water:
+deltaS2 = m2*C*integrate('(1/T)','T',T2,T);// [kJ/K]
+deltaS = deltaS1 + deltaS2;// [kJ/K]
+printf("Change in Entropy is %.3f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.16/Example5_16.sce b/647/CH5/EX5.16/Example5_16.sce new file mode 100755 index 000000000..1136b6157 --- /dev/null +++ b/647/CH5/EX5.16/Example5_16.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 5.16
+// Page: 169
+
+printf("Example: 5.16 - Page: 169\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 12;// [g]
+T1 = 294;// [K]
+T2 = 574;// [K]
+T = 505;// [melting point, K]
+H_fusion = 4.5;// [cal/K]
+C_solid = 0.052;// [cal/g K]
+C_liquid = 0.062;// [cal/g K]
+//*************//
+
+// Entropy Change in heating 12 g of metal from T1 to T
+deltaS1 = m*C_solid*integrate('(1/T)','T',T1,T);// [kJ/K]
+// Entropy Change in fusion of metal:
+deltaS2 = m*H_fusion/T;// [kJ/K]
+// Entropy Change in heating liquid metal from 505 K to 574 K
+deltaS3 = m*C_liquid*integrate('(1/T)','T',T,T2);// [kJ/K]
+deltaS = deltaS1 + deltaS2 + deltaS3;// [kJ/K]
+printf("Change in Entropy is %.3f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.17/Example5_17.sce b/647/CH5/EX5.17/Example5_17.sce new file mode 100755 index 000000000..c22cdc243 --- /dev/null +++ b/647/CH5/EX5.17/Example5_17.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+// Example: 5.17
+// Page: 170
+
+printf("Example: 5.17 - Page: 170\n\n");
+
+// Solution
+
+//*****Data*****//
+deff('[y] = Cp(T)','y = 7.25 + 2.28*10^(-3)*T');
+T1 = 273 + 137;// [K]
+T2 = 273 + 877;// [K]
+//************//
+
+deltaS = integrate('Cp(T)/T','T',T1,T2);// [cal/K]
+printf("Change in Entropy is %.3f cal/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.18/Example5_18.sce b/647/CH5/EX5.18/Example5_18.sce new file mode 100755 index 000000000..082f9d42f --- /dev/null +++ b/647/CH5/EX5.18/Example5_18.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 5.18
+// Page: 170
+
+printf("Example: 5.18 - Page: 170\n\n");
+
+// Solution
+
+//*****Data*****//
+m_iron = 40;// [kg]
+T1 = 625;// [K]
+m_water = 160;// [kg]
+T2 = 276;// [K]
+C_iron = 0.45;// [kJ/kg K]
+C_water = 4.185;// [kJ/kg K]
+//**************//
+
+deff('[y] = f(T)','y = m_iron*C_iron*(T1 - T) - m_water*C_water*(T - T2)');
+T = fsolve(7,f);// [K]
+// Change in Entropy of the iron casting can be estimated as:
+deltaS1 = m_iron*C_iron*log(T/T1);// [kJ/K]
+// Change in Entropy of Water is given by:
+deltaS2 = m_water*C_water*log(T/T2);// [kJ/K]
+deltaS = deltaS1 + deltaS2;// [kJ/K]
+printf("Total Entropy Change is %.2f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.19/Example5_19.sce b/647/CH5/EX5.19/Example5_19.sce new file mode 100755 index 000000000..4dec7015d --- /dev/null +++ b/647/CH5/EX5.19/Example5_19.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 5.19
+// Page: 172
+
+printf("Example: 5.19 - Page: 172\n\n");
+
+// Solution
+
+//*****Data*****//
+Cp = 21;// [J/kmol]
+T1 = 300;// [K]
+T2 = 500;// [K]
+S1 = 150;// [Entropy at T1, J/kmol]
+//*************//
+
+// This is a constant Entropy process. Therefore:
+deltaS = Cp*log(T2/T1);// [J/kmol]
+S2 = S1 + deltaS;// [J/kmol]
+printf("Entropy at 500 K is %.2f J/kmol",S2);
\ No newline at end of file diff --git a/647/CH5/EX5.2/Example5_2.sce b/647/CH5/EX5.2/Example5_2.sce new file mode 100755 index 000000000..71efe00d9 --- /dev/null +++ b/647/CH5/EX5.2/Example5_2.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 5.2
+// Page: 150
+
+printf("Example: 5.2 - Page: 150\n\n");
+
+// Solution
+
+//*****Data*****//
+Th = 810;// [K]
+Tl = 300;// [K]
+//*************//
+
+// Solution (a)
+eta = (Th - Tl)/Th;
+printf("(a) The efficiency of the heat engine is %.1f %%\n",eta*100);
+
+// Solution (b)
+Th = 1366;// [K]
+Tl = 300;// [K]
+eta = (Th - Tl)/Th;
+printf("(b) The efficiency of the heat engine is %.1f %%\n",eta*100);
+
+// Solution (c)
+Th = 810;// [K]
+Tl = 344;// [K]
+eta = (Th - Tl)/Th;
+printf("(c) The efficiency of the heat engine is %.1f %%\n",eta*100);
\ No newline at end of file diff --git a/647/CH5/EX5.20/Example5_20.sce b/647/CH5/EX5.20/Example5_20.sce new file mode 100755 index 000000000..d14cf496d --- /dev/null +++ b/647/CH5/EX5.20/Example5_20.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// Example: 5.20
+// Page: 173
+
+printf("Example: 5.20 - Page: 173\n\n");
+
+// Solution
+
+//*****Data*****//
+T1_oil = 273 + 150;// [K]
+T2_oil = 50 +273;// [K]
+m_water = 4000;// [kg]
+T1_water = 273 + 20;// [K]
+T2_water = 273 + 130;// [K]
+C_water = 4.185;// [kJ/kg K]
+C_oil = 2.5;// [kJ/kg K]
+//***************//
+
+// For oil:
+deltaT_oil = T1_oil - T2_oil;// [K]
+// For water:
+deltaT_water = T2_water - T1_water;// [K]
+// The mass flow rate of oil can be measured by the enthalpy balance over the process:
+m_oil = m_water*C_water*deltaT_water/(deltaT_oil*C_oil);// [kg]
+// Change in the Entropy of oil:
+deltaS_oil = m_oil*C_oil*log(T2_oil/T1_oil);// [kJ/K]
+// Change in Entropy of water:
+deltaS_water = m_water*C_water*log(T2_water/T1_water);// [kJ/K]
+deltaS = deltaS_oil + deltaS_water;// [kJ/K]
+printf("Total Entropy Change is %.2f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.21/Example5_21.sce b/647/CH5/EX5.21/Example5_21.sce new file mode 100755 index 000000000..0a2993e2f --- /dev/null +++ b/647/CH5/EX5.21/Example5_21.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 5.21
+// Page: 174
+
+printf("Example: 5.21 - Page: 174\n\n");
+
+// Solution
+
+//*****Data*****//
+t = 20*60;// [s]
+P = 650;// [W]
+T = 273 + 250;// [K]
+//*************//
+
+Q = P*t/1000;// [kJ]
+deltaS = Q/T;// [kJ/K]
+printf("Change in Entropy is %.2f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.22/Example5_22.sce b/647/CH5/EX5.22/Example5_22.sce new file mode 100755 index 000000000..87d682b7e --- /dev/null +++ b/647/CH5/EX5.22/Example5_22.sce @@ -0,0 +1,33 @@ +clear;
+clc;
+
+// Example: 5.22
+// Page: 174
+
+printf("Example: 5.22 - Page: 174\n\n");
+
+// Solution
+
+//*****Data*****//
+T1 = 400;// [K]
+P1 = 300;// [kPa]
+V1 = 1;// [cubic m]
+V2 =2;// [cubic m]
+R = 8.314;// [kJ/kmol K]
+//**************//
+
+// Since the system is well insulated, there is no scope of transferring heat between system & surrounding.
+deltaQ = 0;// [kJ]
+deltaW = 0;// [kJ]
+// By first law of thermodynamics:
+deltaU =deltaQ - deltaW;// [kJ]
+// As the internal energy of the gas depends only on temperature,
+deltaT = 0;// [K]
+T2 = T1 + deltaT;// [K]
+P2 = (P1*V1/T1)*(T2/V2);// [kPa]
+n = P1*V1/(R*T1);// [kmol]
+deltaS_system = n*R*log(P1/P2);// [kJ/K]
+// Since process is adiabatic:
+deltaS_surrounding = 0;// [kJ/K]
+deltaS = deltaS_system + deltaS_surrounding;// [kJ/K]
+printf("Change in Entropy of the gas is %.4f kJ/K",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.23/Example5_23.sce b/647/CH5/EX5.23/Example5_23.sce new file mode 100755 index 000000000..896937c19 --- /dev/null +++ b/647/CH5/EX5.23/Example5_23.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 5.23
+// Page: 174
+
+printf("Example: 5.23 - Page: 174\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 5.23 on page number 174 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 5.23 on page 174 of the book.")
\ No newline at end of file diff --git a/647/CH5/EX5.24/Example5_24.sce b/647/CH5/EX5.24/Example5_24.sce new file mode 100755 index 000000000..6aa5621b8 --- /dev/null +++ b/647/CH5/EX5.24/Example5_24.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 5.24
+// Page: 182
+
+printf("Example: 5.24 - Page: 182\n\n");
+
+// Solution
+
+// From Example 5.18 (Pg: 170)
+//*****Data*****//
+m_iron = 40;// [kg]
+T1 = 625;// [K]
+m_water = 160;// [kg]
+T2 = 276;// [K]
+C_iron = 0.45;// [kJ/kg K]
+C_water = 4.185;// [kJ/kg K]
+//**************//
+
+deff('[y] = f(T)','y = m_iron*C_iron*(T1 - T) - m_water*C_water*(T - T2)');
+T = fsolve(7,f);// [K]
+// Change in Entropy of the iron casting can be estimated as:
+deltaS1 = m_iron*C_iron*integrate('(1/T)','T',T1,T);// [kJ/K]
+// Change in Entropy of Water is given by:
+deltaS2 = m_water*C_water*integrate('(1/T)','T',T2,T);// [kJ/K]
+deltaS = deltaS1 + deltaS2;// [kJ/K]
+// By Eqn. 5.63:
+W_lost = T2 * deltaS;// [kJ]
+printf("Work lost is %.2f kJ",W_lost);
\ No newline at end of file diff --git a/647/CH5/EX5.25/Example5_25.sce b/647/CH5/EX5.25/Example5_25.sce new file mode 100755 index 000000000..9761340b8 --- /dev/null +++ b/647/CH5/EX5.25/Example5_25.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Example: 5.24
+// Page: 182
+
+printf("Example: 5.24 - Page: 182\n\n");
+
+// Solution
+
+// *****Data******//
+m_oil = 4750;// [kg]
+T1_oil = 515;// [K]
+T2_oil = 315;// [K]
+m_water = 9500;// [kg]
+T1_water = 290;// [K]
+Cp_oil = 3.2;// [kJ/kg K]
+Cp_water = 4.185;// [kJ/kg K]
+//*****************//
+
+// From enthalpy Balance:
+deff('[y] = f(T2_water)','y = m_oil*Cp_oil*(T1_oil - T2_oil) - m_water*Cp_water*(T2_water - T1_water)');
+T2_water = fsolve(7,f);// [K]
+// Change in the Entropy of oil:
+deltaS_oil = m_oil*Cp_oil*integrate('(1/T)','T',T1_oil,T2_oil);// [kJ/K]
+// Change in Entropy of water:
+deltaS_water = m_water*Cp_water*integrate('(1/T)','T',T1_water,T2_water);// [kJ/K]
+deltaS = deltaS_oil + deltaS_water;// [kJ/K]
+printf("Total Entropy Change is %.2f kJ/K\n",deltaS);
+if deltaS > 0
+ printf("Since deltaS is a positive quantity, process is irreversible\n");
+else
+ printf("Since deltaS is a negative quantity, process is reversible\n");
+end
\ No newline at end of file diff --git a/647/CH5/EX5.3/Example5_3.sce b/647/CH5/EX5.3/Example5_3.sce new file mode 100755 index 000000000..4cfc0d395 --- /dev/null +++ b/647/CH5/EX5.3/Example5_3.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 5.3
+// Page: 151
+
+printf("Example: 5.3 - Page: 151\n\n");
+
+// Solution
+
+//*****Data*****//
+Th = 650 + 273;// [K]
+Tl = 30 + 273;// [K]
+Qh = 585;// [kJ/cycle]
+//*************//
+
+// Solution (a)
+// From Eqn. (5.9)
+eta = (Th - Tl)/Th;
+printf("(a) The efficiency of the Carnot engine is %.1f %%\n",eta*100);
+
+// Soluton (b)
+// From the knowledge of the ratio of heat and temperature between the two regions:
+Ql = Qh*Tl/Th;// [kJ]
+printf("(b) Heat released to cold reservoir is %d kJ\n",Ql);
\ No newline at end of file diff --git a/647/CH5/EX5.4/Example5_4.sce b/647/CH5/EX5.4/Example5_4.sce new file mode 100755 index 000000000..bf7a9c7e5 --- /dev/null +++ b/647/CH5/EX5.4/Example5_4.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 5.4
+// Page: 151
+
+printf("Example: 5.4 - Page: 151\n\n");
+
+// Solution
+
+//*****Data*****//
+m = 1;// [kg]
+Tl = 273;// [K]
+Th = 295;// [K]
+Ql = 335;// [kJ/kg]
+//*************//
+
+// Solution (a)
+// The coeffecient of performance of refrigerating machine is:
+// COP = Ql/Wnet = Tl/(Th - Tl)
+Wnet = Ql*(Th - Tl)/Tl;// [kJ]
+printf("Minimum Work requirement is %d kJ\n",round(Wnet));
+
+// Solution (b)
+// Amount of heat released:
+// Wnet = Qh - Ql
+Qh = Wnet + Ql;// [kJ]
+printf("Amount of heat released to the surrounding is %d kJ\n",round(Qh));
\ No newline at end of file diff --git a/647/CH5/EX5.5/Example5_5.sce b/647/CH5/EX5.5/Example5_5.sce new file mode 100755 index 000000000..c77f43fcc --- /dev/null +++ b/647/CH5/EX5.5/Example5_5.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+// Example: 5.5
+// Page: 152
+
+printf("Example: 5.5 - Page: 152\n\n");
+
+// Solution
+
+//*****Data*****//
+Th = 373;// [K]
+Tl = 275;// [K]
+Qh = 50;// [kJ]
+//*************//
+
+// Solution (a)
+// Theral Efficiency of the engine can be given as:
+// eta_HE = Wnet/Qh;
+// Wnet = Qh*COP = Qh*(Th - Tl)/Th;
+Wnet = Qh*(Th - Tl)/Th;// [kJ]
+printf("Minimum Work Required is %.1f kJ\n",Wnet);
+
+// Solution (b)
+eta = (Th - Tl)/Th;
+printf("The efficiency of Heat Engine is %.3f\n",eta);
+
+// Solution (c)
+// Amount of heat released can be calculated as:
+// eta = Net Work Output/Net Work Input;
+// eta = Wnet/Qin;
+// eta = (Qin - Qout)/Qin;
+Qin = Qh;// [kJ]
+Qout = Qin*(1 - eta);
+printf("Amount of Heat released is %.1f kJ\n",Qout);
\ No newline at end of file diff --git a/647/CH5/EX5.6/Example5_6.sce b/647/CH5/EX5.6/Example5_6.sce new file mode 100755 index 000000000..566db3e35 --- /dev/null +++ b/647/CH5/EX5.6/Example5_6.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Example: 5.6
+// Page: 153
+
+printf("Example: 5.6 - Page: 153\n\n");
+
+// Solution
+
+//*****Data*****//
+W = 5;// [hp]
+Q = 7000;// [J/s]
+Th = 400 + 273;// [K]
+Tl = 24 + 273;// [K]
+//*************//
+
+W = 5*745.7;// [W]
+thermal_eta = W/Q;
+theoretical_eta = (Th - Tl)/Th;
+
+if theoretical_eta <= thermal_eta
+ printf("Claim is Valid");
+else
+ printf("Claim is not Valid");
+end
\ No newline at end of file diff --git a/647/CH5/EX5.7/Example5_7.sce b/647/CH5/EX5.7/Example5_7.sce new file mode 100755 index 000000000..c2f4e27d1 --- /dev/null +++ b/647/CH5/EX5.7/Example5_7.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 5.7
+// Page: 162
+
+printf("Example: 5.7 - Page: 162\n\n");
+
+// Solution
+
+//*****Data*****//
+n = 1;// [mol]
+deltaH_fusion = 6500;// [J/mol]
+T_Tr = 273;// [transition temperature, K]
+P = 1;// [atm]
+//************//
+
+// By Eqn. (9.40)
+deltaS_fusion = deltaH_fusion/T_Tr;// [J/mol K]
+printf("Change in Entropy is %.2f J/mol K",deltaS_fusion);
\ No newline at end of file diff --git a/647/CH5/EX5.8/Example5_8.sce b/647/CH5/EX5.8/Example5_8.sce new file mode 100755 index 000000000..cb3908c4f --- /dev/null +++ b/647/CH5/EX5.8/Example5_8.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 5.8
+// Page: 164
+
+printf("Example: 5.8 - Page: 164\n\n");
+
+// Solution
+
+//*****Data*****//
+V1 = 5;// [L]
+V2 = 50;// [L]
+n = 5;// [moles]
+R = 1.987;// [cal/mol K]
+//**************//
+
+// Change in entropy for an isothermal change for an ideal gas:
+// deltaS = n*R*log(P1/P2) = n*R*log(V2/V1)
+deltaS = n*R*log(V2/V1);// [cal/degree]
+printf("Change in Entropy is %.3f eu",deltaS);
\ No newline at end of file diff --git a/647/CH5/EX5.9/Example5_9.sce b/647/CH5/EX5.9/Example5_9.sce new file mode 100755 index 000000000..fb65ac17a --- /dev/null +++ b/647/CH5/EX5.9/Example5_9.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 5.9
+// Page: 164
+
+printf("Example: 5.9 - Page: 164\n\n");
+
+// Solution
+
+//*****Data*****//
+n = 8;// [mol]
+R = 8.314;// [J/mol K]
+T2 = 700;// [K]
+T1 = 350;// [K]
+Cp = (5/2)*R;// [J/mol K]
+//*************//
+
+deltaS = n*Cp*log(T2/T1);// [J/K]
+printf("deltaS is %.2f J/K",deltaS);
\ No newline at end of file diff --git a/647/CH6/EX6.1/Example6_1.sce b/647/CH6/EX6.1/Example6_1.sce new file mode 100755 index 000000000..1ce3732a9 --- /dev/null +++ b/647/CH6/EX6.1/Example6_1.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.1
+// Page: 197
+
+printf("Example: 6.1 - Page: 197\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.1 on page number 197 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.1 on page 197 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.10/Example6_10.sce b/647/CH6/EX6.10/Example6_10.sce new file mode 100755 index 000000000..858ef695e --- /dev/null +++ b/647/CH6/EX6.10/Example6_10.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.10
+// Page: 218
+
+printf("Example: 6.10 - Page: 218\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.10 on page number 218 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.10 on page 218 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.11/Example6_11.sce b/647/CH6/EX6.11/Example6_11.sce new file mode 100755 index 000000000..e64cbe7da --- /dev/null +++ b/647/CH6/EX6.11/Example6_11.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.11
+// Page: 219
+
+printf("Example: 6.11 - Page: 219\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.11 on page number 219 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.11 on page 219 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.12/Example6_12.sce b/647/CH6/EX6.12/Example6_12.sce new file mode 100755 index 000000000..a701edaae --- /dev/null +++ b/647/CH6/EX6.12/Example6_12.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.12
+// Page: 219
+
+printf("Example: 6.12 - Page: 219\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.12 on page number 219 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.12 on page 219 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.13/Example6_13.sce b/647/CH6/EX6.13/Example6_13.sce new file mode 100755 index 000000000..61e23dc64 --- /dev/null +++ b/647/CH6/EX6.13/Example6_13.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.13
+// Page: 220
+
+printf("Example: 6.13 - Page: 220\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 6.13 on page number 220 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 6.13 on page 220 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.14/Example6_14.sce b/647/CH6/EX6.14/Example6_14.sce new file mode 100755 index 000000000..32e335538 --- /dev/null +++ b/647/CH6/EX6.14/Example6_14.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 6.14
+// Page: 221
+
+printf("Example: 6.14 - Page: 221\n\n");
+
+// Solution
+
+// *****Data******//
+alpha = 0.837*10^(-11);// [square m/N]
+beeta = 54.2*10^(-6);// [1/K]
+T = 227 + 273;// [K]
+V = 7.115*10^(-3);// [cubic m/kmol]
+Cp = 26.15;// [J/mol K]
+//*****************//
+
+Cv = Cp - (T*V*beeta^2/alpha)/1000;// [J/mol K]
+// Percentage error if Cp is assumed to Cv.
+err = ((Cp - Cv)/Cp)*100;
+printf("Percentage error is %.2f %%",err);
\ No newline at end of file diff --git a/647/CH6/EX6.15/Example6_15.sce b/647/CH6/EX6.15/Example6_15.sce new file mode 100755 index 000000000..bc07b8d8d --- /dev/null +++ b/647/CH6/EX6.15/Example6_15.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.15
+// Page: 221
+
+printf("Example: 6.15 - Page: 221\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 6.15 on page number 221 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 6.15 on page 221 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.16/Example6_16.sce b/647/CH6/EX6.16/Example6_16.sce new file mode 100755 index 000000000..7fea65697 --- /dev/null +++ b/647/CH6/EX6.16/Example6_16.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.16
+// Page: 222
+
+printf("Example: 6.16 - Page: 222\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 6.16 on page number 222 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 6.16 on page 222 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.17/Example6_17.sce b/647/CH6/EX6.17/Example6_17.sce new file mode 100755 index 000000000..58e24cee9 --- /dev/null +++ b/647/CH6/EX6.17/Example6_17.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.17
+// Page: 222
+
+printf("Example: 6.17 - Page: 222\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 6.17 on page number 222 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 6.17 on page 222 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.18/Example6_18.sce b/647/CH6/EX6.18/Example6_18.sce new file mode 100755 index 000000000..7af1eb72f --- /dev/null +++ b/647/CH6/EX6.18/Example6_18.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.18
+// Page: 223
+
+printf("Example: 6.18 - Page: 223\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 6.18 on page number 223 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 6.18 on page 223 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.19/Example6_19.sce b/647/CH6/EX6.19/Example6_19.sce new file mode 100755 index 000000000..e6cf5d910 --- /dev/null +++ b/647/CH6/EX6.19/Example6_19.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 6.19
+// Page: 227
+
+printf("Example: 6.19 - Page: 227\n\n");
+
+// Solution
+
+// *****Data******//
+a = 3.59;// [square L atm /square mol]
+b = 0.043;// [L/mol]
+R = 0.082;// [J/mol K]
+//***************//
+
+// From Eqn. 6.122:
+Ti = 2*a/(R*b);// [K]
+printf("Inversion of temperature is %.1f K",Ti);
\ No newline at end of file diff --git a/647/CH6/EX6.2/Example6_2.sce b/647/CH6/EX6.2/Example6_2.sce new file mode 100755 index 000000000..45b9b5133 --- /dev/null +++ b/647/CH6/EX6.2/Example6_2.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+// Example: 6.2
+// Page: 205
+
+printf("Example: 6.2 - Page: 205\n\n");
+
+// Solution
+
+// *****Data******//
+density_water = 0.998;// [g/cubic cm]
+density_ice = 0.9168;// [g/cubic cm]
+Hf = 335;// [J/g]
+T = 0 + 273;// [K]
+//*****************//
+
+V_water = 1/density_water;// [cubic cm/g]
+V_ice = 1/density_ice;// [cubic cm/g]
+// From Eqn. 6.56:
+// dP/dT = deltaS/(V2 - V1) = deltaH/(T*(V2 - V1))
+// Substituting these values in Eqn. 6.58
+deltaP_By_deltaT = (Hf/(T*(V_water - V_ice)))*10;// [atm/K]
+deltaT_By_deltaP = 1/deltaP_By_deltaT;// [K/atm]
+if deltaT_By_deltaP > 0
+ printf("Increase in pressure of 1 atm increases the freezing point by %.4f K",abs(deltaT_By_deltaP));
+else
+ printf("Increase in pressure of 1 atm lowers the freezing point by %.4f K",abs(deltaT_By_deltaP));
+end
\ No newline at end of file diff --git a/647/CH6/EX6.20/Example6_20.sce b/647/CH6/EX6.20/Example6_20.sce new file mode 100755 index 000000000..73099d7d2 --- /dev/null +++ b/647/CH6/EX6.20/Example6_20.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.20
+// Page: 227
+
+printf("Example: 6.20 - Page: 227\n\n");
+
+// This problem involves proving a relation in which no mathematics and no calculations are involved.
+// For prove refer to this example 6.20 on page number 227 of the book.
+
+printf(" This problem involves proving a relation in which no mathematics and no calculations are involved.\n\n");
+printf(" For prove refer to this example 6.20 on page 227 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.21/Example6_21.sce b/647/CH6/EX6.21/Example6_21.sce Binary files differnew file mode 100755 index 000000000..53aceff60 --- /dev/null +++ b/647/CH6/EX6.21/Example6_21.sce diff --git a/647/CH6/EX6.22/Example6_22.sce b/647/CH6/EX6.22/Example6_22.sce Binary files differnew file mode 100755 index 000000000..d82ae8d13 --- /dev/null +++ b/647/CH6/EX6.22/Example6_22.sce diff --git a/647/CH6/EX6.23/Example6_23.sce b/647/CH6/EX6.23/Example6_23.sce new file mode 100755 index 000000000..16cc2e459 --- /dev/null +++ b/647/CH6/EX6.23/Example6_23.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+// Example: 6.23
+// Page: 239
+
+printf("Example: 6.23 - Page: 239\n\n");
+
+// Solution
+
+// *****Data******//
+T = 298;// [K]
+P = 10*10^5;// [Pa]
+Tc = 126.2;// [K]
+Pc = 34*10^5;// [bar]
+R = 8.314;// [J/mol K]
+//****************//
+
+a = 27*R^2*Tc^2/(64*Pc);// [Pa.m^6/square mol]
+b = R*Tc/(8*Pc);// [cubic m/mol]
+V = 2.425*10^(-3);// [cubic m/mol]
+// From Eqn. 6.173:
+Sr = R*log(P*(V - b)/(R*T));// [J/mol K]
+printf("Residual Entropy is %.4f J/mol K\n",Sr);
+// From Eqn. 6.174:
+Hr = P*V - R*T - (a/V);// [J/mol]
+printf("Residual Enthalpy is %.4f J/mol K\n",Hr);
+Ur = -(a/V);// [J/mol]
+printf("Residual Internal Energy is %.4f J/mol K\n",Ur);
\ No newline at end of file diff --git a/647/CH6/EX6.24/Example6_24.sce b/647/CH6/EX6.24/Example6_24.sce new file mode 100755 index 000000000..c454da6aa --- /dev/null +++ b/647/CH6/EX6.24/Example6_24.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 6.24
+// Page: 244
+
+printf("Example: 6.24 - Page: 244\n\n");
+
+// Solution
+
+// *****Data******//
+B = -4.28*10^(-4);// [cubic m/mol]
+P = 15*10^5;// [Pa]
+T = 273 + 87;// [K]
+R = 8.314;// [J/atm K]
+//****************//
+
+// Z = 1 + (B*P/(R*T))
+// (Z - 1)/P = B/(R*T)
+// From Eqn. 6.192 (b)
+// ln(f/P) = integral('(Z - 1)/P','P',0,P) = B*P/(R*T)
+f = P*exp(B*P/(R*T));// [Pa]
+printf("Fugacity of iso-butane is %.2f atm",f/10^5);
\ No newline at end of file diff --git a/647/CH6/EX6.3/Example6_3.sce b/647/CH6/EX6.3/Example6_3.sce new file mode 100755 index 000000000..cc3d20107 --- /dev/null +++ b/647/CH6/EX6.3/Example6_3.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 6.3
+// Page: 206
+
+printf("Example: 6.3 - Page: 206\n\n");
+
+// Solution
+
+// *****Data******//
+P1 = 361.3;// [kPa]
+T1 = 140 + 273;// [K]
+P2 = 617.8;// [kPa]
+T2 = 160 + 273;// [K]
+T = 150 + 273;// [K]
+Vg = 0.3917;// [cubic m/kg]
+//****************//
+
+// From Eqn. 6.56
+// dP/dT = deltaH/(T*(Vg - V1)) = deltaH/(T*Vg)
+deltaP = P2 - P1;// [kPa]
+deltaT = T2 - T1;// [K]
+deltaH = T*Vg*deltaP/deltaT;// [kJ/kg]
+printf("Enthalpy of Vaporisation is %d kJ/kg\n",round(deltaH));
\ No newline at end of file diff --git a/647/CH6/EX6.4/Example6_4.sce b/647/CH6/EX6.4/Example6_4.sce new file mode 100755 index 000000000..78d472055 --- /dev/null +++ b/647/CH6/EX6.4/Example6_4.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 6.4
+// Page: 206
+
+printf("Example: 6.4 - Page: 206\n\n");
+
+// Solution
+
+// *****Data******//
+T1 = -40 + 273;// [K]
+T2 = -45 + 273;// [K]
+P1 = 51.25;// [kPa]
+R = 0.0815;// [kJ/kg K]
+Hv = 225.86;// [kJ/kg]
+//****************//
+
+// From Eqn. 6.61:
+P2 = P1*exp((Hv/R)*((1/T1) - (1/T2)));// [kPa]
+printf("Saturation pressure of the refrigerant is %.2f kPa",P2);
\ No newline at end of file diff --git a/647/CH6/EX6.5/Example6_5.sce b/647/CH6/EX6.5/Example6_5.sce new file mode 100755 index 000000000..53136b998 --- /dev/null +++ b/647/CH6/EX6.5/Example6_5.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 6.5
+// Page: 206
+
+printf("Example: 6.5 - Page: 206\n\n");
+
+// Solution
+
+// *****Data******//
+Tb = -103.9 + 273;// [K]
+deff('[P] = f1(T)','P = 10^(-(834.13/T) + 1.75*log10(T) - 8.375*10^(-3)*T + 5.324)');
+R = 8.314;// [J/mol K]
+//***************//
+
+// From Eqn. 6.60, we get:
+// d(ln(P))/dT = deltaH/(R*T^2)
+deff('[P] = f2(T)','P = exp(2.303*log10(f1(T)))');
+// Differentiating it with respect to T
+// d(ln(P))/dT = (834.13*2.303/Tb^2 + 1.75/Tb - 2.303*8.375*10^(-3))
+deltaH = R*Tb^2*(834.13*2.303/Tb^2 + 1.75/Tb - 2.303*8.375*10^(-3))/1000;// [kJ/mol]
+printf("The enthalpy of vaporisation is %.2f kJ/mol\n",deltaH);
\ No newline at end of file diff --git a/647/CH6/EX6.6/Example6_6.sce b/647/CH6/EX6.6/Example6_6.sce new file mode 100755 index 000000000..0352ce575 --- /dev/null +++ b/647/CH6/EX6.6/Example6_6.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.6
+// Page: 214
+
+printf("Example: 6.6 - Page: 214\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.6 on page number 214 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.6 on page 214 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.7/Example6_7.sce b/647/CH6/EX6.7/Example6_7.sce new file mode 100755 index 000000000..04d20c26f --- /dev/null +++ b/647/CH6/EX6.7/Example6_7.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.7
+// Page: 215
+
+printf("Example: 6.7 - Page: 215\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.7 on page number 215 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.7 on page 215 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.8/Example6_8.sce b/647/CH6/EX6.8/Example6_8.sce new file mode 100755 index 000000000..b7853227b --- /dev/null +++ b/647/CH6/EX6.8/Example6_8.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.8
+// Page: 217
+
+printf("Example: 6.8 - Page: 217\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.8 on page number 217 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.8 on page 217 of the book.");
\ No newline at end of file diff --git a/647/CH6/EX6.9/Example6_9.sce b/647/CH6/EX6.9/Example6_9.sce new file mode 100755 index 000000000..afb47b082 --- /dev/null +++ b/647/CH6/EX6.9/Example6_9.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 6.9
+// Page: 217
+
+printf("Example: 6.9 - Page: 217\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 6.9 on page number 217 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 6.9 on page 217 of the book.");
\ No newline at end of file diff --git a/647/CH7/EX7.1/Example7_1.sce b/647/CH7/EX7.1/Example7_1.sce new file mode 100755 index 000000000..56a70f23a --- /dev/null +++ b/647/CH7/EX7.1/Example7_1.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+
+// Example: 7.1
+// Page: 256
+
+printf("Example: 7.1 - Page: 256\n\n");
+
+// Solution
+
+// *****Data******//
+d1 = 0.15;// [inlet dia, m]
+d2 = 0.20;// [outlet dia, m]
+U1 = 7;// [inlet velocity, m/s]
+//****************//
+
+// From Fig. 7.2 (Pg 256)
+// At the inlet:
+A1 = (%pi/4)*d1^2;// [square m]
+// At the outlet:
+A2 = (%pi/4)*d2^2;// [square m]
+Q = A1*U1;// [cubic m/s]
+printf("Flow rate is %.4f m/s\n",Q);
+// Using Continuity Eqn.
+// density1*U1*A1 = Density2*U2*A2
+// For water: Density1 = Density2. Therefore:
+U2 = A1*U1/A2;
+printf("Velocity of water at the outlet is %.3f m/s",U2);
\ No newline at end of file diff --git a/647/CH7/EX7.10/Example7_10.sce b/647/CH7/EX7.10/Example7_10.sce new file mode 100755 index 000000000..386fb15bd --- /dev/null +++ b/647/CH7/EX7.10/Example7_10.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 7.10
+// Page: 280
+
+printf("Example: 7.10 - Page: 280\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 800;// [kPa]
+T1 = 773;// [K]
+H1 = 3480;// [kJ/kg]
+P2 = 100;// [kPa]
+T2 = 573;// [K]
+H2 = 3074;// [kJ/kg]
+//***************//
+
+// Solution (a)
+// Velocity of the fluid exiting the nozzle:
+// U2 = sqrt(U1^2 + 2*(H1 - H2))
+// Neglecting initial velocity:
+U2 = sqrt(2*(H1 - H2)*1000);// [m/s]
+printf("(a) Final Velocity is %.2f m/s\n",U2);
+
+// Solution (b)
+U1 = 40;// [m/s]
+U2 = sqrt((U1^2 + 2*(H1 - H2))*1000);// [m/s]
+printf("(b) Final Velocity is %.2f m/s\n",U2);
\ No newline at end of file diff --git a/647/CH7/EX7.11/Example7_11.sce b/647/CH7/EX7.11/Example7_11.sce new file mode 100755 index 000000000..564f93cfd --- /dev/null +++ b/647/CH7/EX7.11/Example7_11.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+
+// Example: 7.11
+// Page: 281
+
+printf("Example: 7.11 - Page: 281\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 100;// [kPa]
+T1 = 200;// [OC]
+U1 = 190;// [m/s]
+A1 = 2000/10^4;// [square m]
+U2 = 70;// [m/s]
+P2 = 200;// [kPa]
+Qdot = 100;// [kW]
+V1 = 2.172;// [cubic m/kg]
+H1 = 2875.3;// [kJ/kg]
+//***************//
+
+// Solution (a)
+mdot = U1*A1/V1;// [kg/s]
+printf("Mass flow rate of the steam is %.2f kg/s\n",mdot);
+
+// Solution (b)
+// Amount of heat transferred to the surrounding per unit steam:
+Q = Qdot/mdot;// [kJ/kg]
+// The Enthalpy at the diffuser outlet can be estimated as:
+H2 = Q + H1 + (U1^2 - U2^2)/2;// [kJ/kg]
+// From the steam table:
+T2 = 393.38;// [K]
+V2 = 1.123;// [cubic m/kg]
+printf("The temperature of the steam leaving the outlet is %.2f K\n",T2);
+
+// Solution (c)
+A2 = V2*mdot/U2;// [square m]
+printf("Area of diffuser outlet is %.2f square m\n",A2);
\ No newline at end of file diff --git a/647/CH7/EX7.2/Example7_2.sce b/647/CH7/EX7.2/Example7_2.sce new file mode 100755 index 000000000..4e5ddf68c --- /dev/null +++ b/647/CH7/EX7.2/Example7_2.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// Example: 7.2
+// Page: 257
+
+printf("Example: 7.2 - Page: 257\n\n");
+
+// Solution
+
+//*****Data******//
+d1 = 0.2;// [m]
+d2 = 0.15;// [m]
+d3 = 0.1;// [m]
+U1 = 3;// [m/s]
+U2 = 2.5;// [m/s]
+//**************//
+
+// From Fig. 7.3 (Pg: 257)
+// For pipe I:
+A1 = (%pi/4)*d1^2;// [square m]
+Q1 = A1*U1;// [cubic m/s]
+// For pipe II:
+A2 = (%pi/4)*d2^2;// [square m]
+Q2 = A2*U2;// [cubic m/s]
+// For pipe III:
+A3 = (%pi/4)*d3^2;// [square m]
+// From continuity Eqn.:
+Q3 = Q1 - Q2;// [cubic m/s]
+U3 = Q3/A3;// [m/s]
+printf("Discharge through the 10 cm pipe is %.4f cubic m/sec\n",Q1);
+printf("Average velocity in the 15 cm pipe is %.2f m/s",U3);
\ No newline at end of file diff --git a/647/CH7/EX7.3/Example7_3.sce b/647/CH7/EX7.3/Example7_3.sce new file mode 100755 index 000000000..863ea29be --- /dev/null +++ b/647/CH7/EX7.3/Example7_3.sce @@ -0,0 +1,29 @@ +clear;
+clc;
+
+// Example: 7.3
+// Page: 262
+
+printf("Example: 7.3 - Page: 262\n\n");
+
+// Solution
+
+//*****Data******//
+d1 = 0.3;// [m]
+d2 = 0715;//[m]
+Q = 40/1000;// [cubic m/s]
+Z1 = 8;// [m]
+Z2 = 6;// [m]
+P1 = 5*10^5;// [Pa]
+density = 1000;// [kg/cubic m]
+g = 9.81;// [m/square s]
+//*************//
+
+// From Fig. 7.3 (Pg: 262)
+A1 = (%pi/4)*d1^2;// [square m]
+A2 = (%pi/4)*d2^2;// [square m]
+U1 = Q/A1;// [m/s]
+U2 = Q/A2;// [m/s]
+// Applying Bernoulli's equations at sections 1 & 2:
+P2 = ((U1^2/(2*g) + Z1 + P1/(density*g)) - (U2^2/(2*g) + Z2))*(density*g);// [Pa]
+printf("Pressure at section 2 is %.2f bar",P2/10^5);
\ No newline at end of file diff --git a/647/CH7/EX7.4/Example7_4.sce b/647/CH7/EX7.4/Example7_4.sce new file mode 100755 index 000000000..ad744d058 --- /dev/null +++ b/647/CH7/EX7.4/Example7_4.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+// Example: 7.4
+// Page: 268
+
+printf("Example: 7.4 - Page: 268\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 100;// [kPa]
+T1 = 320;// [K]
+P2 = 600;// [kPa]
+T2 = 430;// [K]
+m_dot = 0.03;// [kg/s]
+Qout = 15;// [kJ/kg]
+//*************//
+
+// The energy balance around the compressor:
+// dE_System/dt = Ein - Eout
+// Since it is a steady state process:
+// dE_Sysytem/dt = 0
+// Ein = Eout
+// Win + m_dot*H1 = Qout + m_dot*H2
+// Since, Qout = Qout/m
+// Win = m_dot*(Qout + (H2 - H1))
+// From enthalpy chart of air:
+H1 = 320.20;// [Enthalpy of air at 320 K, kJ/kg]
+H2 = 431.43;// [Enthalpy of air at 430 K, kJ/kg]
+Win = m_dot*(Qout + (H2 - H1));// [kW]
+printf("Power Requirement of the compressor is %.2f kW",Win);
\ No newline at end of file diff --git a/647/CH7/EX7.5/Example7_5.sce b/647/CH7/EX7.5/Example7_5.sce new file mode 100755 index 000000000..74d1a60fb --- /dev/null +++ b/647/CH7/EX7.5/Example7_5.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+
+// Example: 7.5
+// Page: 269
+
+printf("Example: 7.5 - Page: 269\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 100;// [kPa]
+T1 = 250;// [K]
+Q = 0.1;// [cubic m/s]
+P2 = 500;// [kPa]
+M = 44;// [g/mol]
+R = 8.314;// [J/mol K]
+//****************//
+
+// Solution (a)
+// Work done by reversible adiabatic compression, gama = 1.4;
+gama = 1.4;
+T2 = T1*(P2/P1)^((gama - 1)/gama);// [K]
+Wad = (gama*R/(gama - 1))*(T1 - T2);// [J/mol]
+Wad = Wad/M;// [J/g]
+printf("Work done by reversible adiabatic compression when gama = 1.4 is %.2f J/g\n",Wad);
+
+// Solution (b)
+// Work done by isothermal compression:
+Wiso = - (R*T1)*log(P2/P1);// [J/mol]
+Wiso = Wiso/M;// [J/g]
+printf("Work done by isothermal compression is %.2f J/g\n",Wiso);
+
+// Solution (c)
+// Work done in single stage compression, gama = 1.3:
+gama = 1.3;
+V1 = Q;// [cubic m]
+Wsingle_stage = (gama*P1*V1/(gama - 1))*(1-(P2/P1)^((gama - 1)/gama));// [kW]
+printf("Work done in single stage compression is %.2f kW",Wsingle_stage);
\ No newline at end of file diff --git a/647/CH7/EX7.6/Example7_6.sce b/647/CH7/EX7.6/Example7_6.sce Binary files differnew file mode 100755 index 000000000..a63fcd742 --- /dev/null +++ b/647/CH7/EX7.6/Example7_6.sce diff --git a/647/CH7/EX7.7/Example7_7.sce b/647/CH7/EX7.7/Example7_7.sce new file mode 100755 index 000000000..691bb18a0 --- /dev/null +++ b/647/CH7/EX7.7/Example7_7.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 7.7
+// Page: 274
+
+printf("Example: 7.7 - Page: 274\n\n");
+
+// Solution
+
+//*****Data******//
+T_steam1 = 50;// [OC]
+T_steam2 = 30;// [OC]
+msteam_dot = 10;// [kg/min]
+T_water1 = 15;// [OC]
+T_water2 = 25;// [OC]
+//***************//
+
+// Solution (a)
+// From the Stem Table:
+H1 = 2645.9;// [kJ/kg, At 50 OC]
+H2 = 768.2;// [kJ/kg, At 30 OC]
+H3 = 62.982;// [kJ/kg, At 15 OC]
+H4 = 104.83;// [kJ/kg, At 25 OC]
+// The mass & Energy balance of the above flow gives:
+mwater_dot = msteam_dot*(H1 - H2)/(H4 - H3);// [kg/min]
+printf("The mass flow rate of water is %.2f kg/min\n",mwater_dot);
+
+// Solution (b)
+Qdot = mwater_dot*(H4 - H3);// [kJ/min]
+printf("The rate of heat transfer is %.2f kJ/min",Qdot);
\ No newline at end of file diff --git a/647/CH7/EX7.8/Example7_8.sce b/647/CH7/EX7.8/Example7_8.sce new file mode 100755 index 000000000..0dc2298f1 --- /dev/null +++ b/647/CH7/EX7.8/Example7_8.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 7.8
+// Page: 279
+
+printf("Example: 7.8 - Page: 279\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 500;// [kPa]
+T1 = 623;// [K]
+mdot = 12;// [kg/s]
+P2 = 500;// [kPa]
+T2 = 523;// [K]
+Qdot = -120;// [kW]
+H1 = 3168;// [kJ/kg]
+H2 = 2976;// [kJ/kg]
+//************//
+
+Q = Qdot/mdot;// [kJ/kg]
+// By energy balance:
+// (deltaU^2/2) + g*deltaZ + deltaH = Q - Ws
+// Considering negligible change in P.E., deltaZ = 0 & Ws = 0.
+// (U2^2 - U1^2)/2 + deltaH = Q
+deltaH = H2 - H1;// [kJ/kg]
+// Neglecting inlet velocity.
+U2 = sqrt(2*(Q - deltaH)*1000);// [m/s]
+printf("Outlrt velocity is %.1f m/s\n",U2);
\ No newline at end of file diff --git a/647/CH7/EX7.9/Example7_9.sce b/647/CH7/EX7.9/Example7_9.sce new file mode 100755 index 000000000..11277cc48 --- /dev/null +++ b/647/CH7/EX7.9/Example7_9.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+
+// Example: 7.9
+// Page: 279
+
+printf("Example: 7.9 - Page: 279\n\n");
+
+// Solution
+
+//*****Data******//
+Pin = 1000;// [kPa]
+Tin = 600;// [K]
+Uin = 50;// [m/s]
+gama = 1.4;
+M = 17;// [g/mol]
+R = 8314;// [kJ/mol K]
+MachNumber = 2;
+//***************//
+
+// Solution (i)
+// Using Eqn. (7.36):
+Critical_Ratio = (2/(gama + 1))^(gama/(gama - 1));
+printf("Critical Ratio is %.2f\n",Critical_Ratio);
+
+// Solution (ii)
+PV_in = R*Tin/M;// [square m]
+Uthroat = sqrt(Uin^2 + (2*gama*PV_in/(gama - 1))*(1-(Critical_Ratio)^((gama - 1)/gama)));// [m/s]
+Uout = MachNumber*Uthroat;// [m/s]
+printf("The discharge velocity is %.2f m/s",Uout);
\ No newline at end of file diff --git a/647/CH8/EX8.1/Example8_1.sce b/647/CH8/EX8.1/Example8_1.sce new file mode 100755 index 000000000..516ac7fc3 --- /dev/null +++ b/647/CH8/EX8.1/Example8_1.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 8.1
+// Page: 297
+
+printf("Example: 8.1 - Page: 297\n\n");
+
+// Solution
+
+//*****Data******//
+Tl = 273 - 4;// [K]
+Th = 273 + 30;// [K]
+Ql = 30;// [kW]
+//*************//
+
+// Solution (a)
+COP = Tl/(Th - Tl);
+printf("The coeffecient of performance of the cycle is %.2f\n",COP);
+
+// Solution (b)
+Wnet = Ql/COP;// [kW]
+printf("The power required is %.2f kW\n",Wnet);
+
+// Solution (c)
+Qh = Wnet + Ql;// [kW]
+printf("The rate of heat rejection in the room is %.2f kW",Qh);
\ No newline at end of file diff --git a/647/CH8/EX8.10/Example8_10.sce b/647/CH8/EX8.10/Example8_10.sce new file mode 100755 index 000000000..bce25e4d2 --- /dev/null +++ b/647/CH8/EX8.10/Example8_10.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+// Example: 8.10
+// Page: 313
+
+printf("Example: 8.10 - Page: 313\n\n");
+
+// Solution
+
+//*****Data******//
+Q = 5;// [tons]
+T1 = 253;// [Temperature of the working fluid leaving the evaporator, K]
+T2 = 303;// [Temperature of the working fluid leaving the evaporator, K]
+T3 = 303;// [K]
+Pressure_Ratio = 4;
+C = 1.008;// [kJ/kg]
+gama = 1.4;
+//**************//
+
+// Solution (a)
+T2 = T1*((Pressure_Ratio)^((gama - 1)/gama));// [K]
+T2 = T1*(Pressure_Ratio)^((gama - 1)/gama);// [K]
+T4 = T3/((Pressure_Ratio)^((gama - 1)/gama));// [K]
+COP = T1/(T2 - T1);
+printf("COP of Air Refrigeration System is %.2f\n",COP);
+
+// Solution (b)
+mdot = Q*12660/(C*(T1 - T4));// [kg/h]
+printf("Mass flow rate of the refrigerant is %.2f kg/h\n",mdot);
+
+// Solution (c)
+Wcompression = mdot*C*(T2 - T3);// [kJ/h]
+printf("The work of Compression is %.2f kW\n",Wcompression/3600);
+
+// Solution (d)
+Wexpansion = mdot*C*(T1 - T4);// [kJ/h]
+printf("The work of expansion is %.2f kW\n",Wexpansion/3600);
+
+// Solution (e)
+Wnet = Wcompression - Wexpansion;// [kJ/h]
+printf("Net work of the system is %.2f kW\n",Wnet/3600);
\ No newline at end of file diff --git a/647/CH8/EX8.2/Example8_2.sce b/647/CH8/EX8.2/Example8_2.sce new file mode 100755 index 000000000..8f028888f --- /dev/null +++ b/647/CH8/EX8.2/Example8_2.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 8.2
+// Page: 298
+
+printf("Example: 8.2 - Page: 298\n\n");
+
+// Solution
+
+//*****Data******//
+Tl = -10 + 273;// [K]
+Th = 45 + 273;// [K]
+Ql = 1;// [ton]
+//*************//
+
+// Solution (a)
+COP = Tl/(Th - Tl);
+Wnet = Ql*3.516/COP;// [kW]
+printf("The power consumption is %.3f kW\n",Wnet);
+
+// Solution (b)
+Tl = -20 + 273;// [K]
+Th = 45 + 273;// [K]
+COP = Tl/(Th - Tl);
+Ql = Wnet*COP;// [kW]
+printf("Cooling Effect produced is %.3f kW\n",Ql);
\ No newline at end of file diff --git a/647/CH8/EX8.3/Example8_3.sce b/647/CH8/EX8.3/Example8_3.sce new file mode 100755 index 000000000..865dbb4d4 --- /dev/null +++ b/647/CH8/EX8.3/Example8_3.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 8.3
+// Page: 298
+
+printf("Example: 8.3 - Page: 298\n\n");
+
+// Solution
+
+// From Example 8.2:
+
+// For refrigerated space:
+// Wnet = Ql/4.78 = 0.209*Ql;
+
+// For freezer box.
+// Wnet = Ql/3.89 = 0.257*Ql
+
+percent = ((0.257 - 0.209)/0.209)*100;
+printf("Increase in percentage of work output is %.2f %%",percent);
\ No newline at end of file diff --git a/647/CH8/EX8.4/Example8_4.sce b/647/CH8/EX8.4/Example8_4.sce new file mode 100755 index 000000000..36ade8c4c --- /dev/null +++ b/647/CH8/EX8.4/Example8_4.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+// Example: 8.4
+// Page: 299
+
+printf("Example: 8.4 - Page: 299\n\n");
+
+// Solution
+
+//*****Data******//
+Th = 273 + 24;// [K]
+Tl = 0 + 273;// [K]
+Qh = 25;// [kW]
+//*************//
+
+COP = Th/(Th - Tl);
+Wnet = Qh/COP;// [kW]
+printf("Coeffecient of performance of Carnot Heat Pump is %.2f\n",COP);
+printf("Power input can be estimated as %.2f kW\n",Wnet);
\ No newline at end of file diff --git a/647/CH8/EX8.5/Example8_5.sce b/647/CH8/EX8.5/Example8_5.sce new file mode 100755 index 000000000..64f7c310c --- /dev/null +++ b/647/CH8/EX8.5/Example8_5.sce @@ -0,0 +1,19 @@ +clear;
+clc;
+
+// Example: 8.5
+// Page: 299
+
+printf("Example: 8.5 - Page: 299\n\n");
+
+// Solution
+
+//*****Data******//
+Tl = -2 + 273;// [K]
+Th = 20 + 273;// [K]
+Qh = 80000;// [kJ/h]
+//*************//
+
+Ql = Qh*Tl/Th;// [kJ/h]
+Wnet = Qh - Ql;// [kJ/h]
+printf("Minimum Power input required is %.3f kW\n",Wnet/3600);
\ No newline at end of file diff --git a/647/CH8/EX8.6/Example8_6.sce b/647/CH8/EX8.6/Example8_6.sce new file mode 100755 index 000000000..80e6e365c --- /dev/null +++ b/647/CH8/EX8.6/Example8_6.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+// Example: 8.6
+// Page: 303
+
+printf("Example: 8.6 - Page: 303\n\n");
+
+// Solution
+
+//*****Data******//
+Tl = 273;// [K]
+Th = 313;// [K]
+H1 = 187;// [Enthalpy of saturated vapour at 273 K, kJ/kg]
+H3 = 74;// [Enthalpy of saturated liquid at 313 K,kJ/kg]
+H4 = H3;// [kJ/kg]
+H2 = 204;// [Enthalpy of Supersaturated Vapour at 273 K, kJ/kg]
+//****************//
+
+// Solution (i)
+// COP = Ql/Wnet;
+COP = ((H1 - H4)/(H2 - H1));
+printf("Enthalpy of saturated vapour is %.2f\n",COP);
+
+// Solution (ii)
+Ref_Effect = H1 - H4;// [kJ/kg]
+printf("Refrigerating Effect is %d kJ/kg\n",Ref_Effect);
+
+// Solution (iii)
+COP = Tl/(Th - Tl);
+printf("The COP of an ideal Carnot refrigerator is %.2f\n",COP);
+
+// Solution (iv)
+W = H2 - H1;// [kJ/kg]
+printf("Work done by the compression is %.2f kJ/kg\n",W);
\ No newline at end of file diff --git a/647/CH8/EX8.7/Example8_7.sce b/647/CH8/EX8.7/Example8_7.sce new file mode 100755 index 000000000..9ccf891dd --- /dev/null +++ b/647/CH8/EX8.7/Example8_7.sce @@ -0,0 +1,35 @@ +clear;
+clc;
+
+// Example: 8.7
+// Page: 304
+
+printf("Example: 8.7 - Page: 304\n\n");
+
+// Solution
+
+//*****Data******//
+P1 = 0.18;// [MPa]
+T1 = -10 + 273;// [K]
+mdot = 0.06;// [kg/s]
+P2 = 1;// [MPa]
+T2 = 45 + 273;// [K]
+T = 273 + 29;// [K]
+P = 0.75;// [MPa]
+H1 = 245.16;// [Enthalpy of superheated vapour at -10 OC & 0.18 MPa, kJ/kg]
+H2 = 277.2;// [Enthalpy of superheated vapour at 45 OC & 1 MPa, kJ/kg]
+H3 = 92.22;// [Enthalpy of saturated liquid at 29 OC & 0.75 MPa, kJ/kg]
+H4 = H3;// [kJ/kg]
+//*************//
+
+// Solution (a)
+Ql = mdot*(H1 - H4);// [kW]
+printf("Amount of heat removed from cold space is %.2f kW\n",Ql);
+
+// Solution (b)
+Wnet = mdot*(H2 - H1);// [kW]
+printf("THe power input required is %.2f kW\n",Wnet);
+
+// Solution (c)
+COP = Ql/Wnet;
+printf("COP of refrigeration of cycle is %.2f\n",COP);
\ No newline at end of file diff --git a/647/CH8/EX8.8/Example8_8.sce b/647/CH8/EX8.8/Example8_8.sce new file mode 100755 index 000000000..da9b430b3 --- /dev/null +++ b/647/CH8/EX8.8/Example8_8.sce @@ -0,0 +1,41 @@ +clear;
+clc;
+
+// Example: 8.8
+// Page: 305
+
+printf("Example: 8.8 - Page: 305\n\n");
+
+// Solution
+
+//*****Data******//
+Ql = 5;// [tons]
+Tl = -10 + 273;// [K]
+Th = 35 + 273;// [K]
+eta = 0.85;
+H1 = 183.2;// [Enthalpy of saturated vapour at 263 K, kJ/kg]
+H2 = 208.3;// [Enthalpy of superheated vapour, kJ/kg]
+H3 = 69.5;// [Enthalpy of saturated vapour at 308 K, kJ/kg]
+H4 = H3;// [kJ/kg]
+//***************//
+
+// Solution (a)
+// Mass flow rate:
+Ql = Ql*3.516;// [kW]
+mdot = Ql/(H1 - H4);// [kW]
+printf("Mass flow rate of the refrigerant is %.4f kg/s\n",mdot);
+
+// Solution (b)
+W = H2 - H1;// [kJ/kg]
+Wnet = W*mdot/eta;// [kW]
+printf("Power consumption in the compression is %.2f kW\n",Wnet);
+
+// Solution (c)
+Qh = Ql + Wnet;// [kW]
+printf("The amount of heat rejected in the condenser is %.2f kW\n",Qh);
+
+// Solution (d)
+COP_VapourCompression = (H1 - H4)/(H2 - H1);
+COP_Carnot = Tl/(Th - Tl);
+COP_relative = COP_VapourCompression/COP_Carnot;
+printf("Relative COP is %.2f\n",COP_relative);
\ No newline at end of file diff --git a/647/CH8/EX8.9/Example8_9.sce b/647/CH8/EX8.9/Example8_9.sce new file mode 100755 index 000000000..37914fbff --- /dev/null +++ b/647/CH8/EX8.9/Example8_9.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 8.9
+// Page: 308
+
+printf("Example: 8.9 - Page: 308\n\n");
+
+// Solution
+
+//*****Data******//
+Th = 273 + 125;// [K]
+Tl = 273 - 5;// [K]
+Ts = 273 + 28;// [K]
+COP = 2;
+//*************//
+
+COP_absorption = (Tl/(Ts - Tl))*((Th - Ts)/Th);
+if (COP - 0.1) < COP_absorption | (COP + 0.1) > COP_absorption
+ printf("Claim is Valid and reasonable");
+else
+ printf("Claim is not Valid");
+end
\ No newline at end of file diff --git a/647/CH9/EX9.1/Example9_1.sce b/647/CH9/EX9.1/Example9_1.sce new file mode 100755 index 000000000..cf49ae51d --- /dev/null +++ b/647/CH9/EX9.1/Example9_1.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 9.1
+// Page: 338
+
+printf("Example: 9.1 - Page: 338\n\n");
+
+// Solution
+
+//*****Data******//
+V1_bar = 52.37*10^(-6);// [partial molar volume of ethanol, cubic m/mol]
+y1 = 0.5;// [mole fraction of ethanol]
+Density = 800.21;// [kg/cubic m]
+M1 = 46*10^(-3);// //[Molecular wt. of ethanol,kg/mol]
+M2 = 18*10^(-3);// [Molecular wt. of water,kg/cmol]
+//*************//
+
+y2 = 1 - y1;// [mole fraction of water]
+M = y1*M1 + y2*M2;// [Molecular wt. of mixture, kg/mol]
+V = M/Density;// [Volume of mixture, cubic m/mol]
+// From Eqn. 9.9:
+V2_bar = (V - y1*V1_bar)/y2;// [partial molar volume of water, cubic m/mol]
+printf("Partial molar volume of water is %.2e cubic m/mol\n",V2_bar);
\ No newline at end of file diff --git a/647/CH9/EX9.10/Example9_10.sce b/647/CH9/EX9.10/Example9_10.sce new file mode 100755 index 000000000..4a99ac9fd --- /dev/null +++ b/647/CH9/EX9.10/Example9_10.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+// Example: 9.10
+// Page: 354
+
+printf("Example: 9.10 - Page: 354\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.3;// [mole fraction of component 1 in the mixture]
+x2 = 0.7;// [mole fraction of component 2 in the mixture]
+phi1 = 0.7;// [fugacity coeffecient of component 1 in the mixture]
+phi2 = 0.85;// [fugacity coeffecient of component 2 in the mixture]
+P = 50;// [bar]
+T = 273 + 100;// [K]
+//*************//
+
+phi = exp(x1*log(phi1) + x2*log(phi2));// [fugacity coeffecient of the mixture]
+f = phi*P;// [bar]
+printf("Fugacity of the gaseous mixture is %.3f bar",f);
\ No newline at end of file diff --git a/647/CH9/EX9.11/Example9_11.sce b/647/CH9/EX9.11/Example9_11.sce new file mode 100755 index 000000000..dfb750d00 --- /dev/null +++ b/647/CH9/EX9.11/Example9_11.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 9.11
+// Page: 354
+
+printf("Example: 9.11 - Page: 354\n\n");
+
+// Solution
+
+//*****Data******//
+x1 = 0.3;// [mole fraction of hydrogen in the mixture]
+x2 = 0.25;// [mole fraction of nitrogen in the mixture]
+x3 = 0.45;// [mole fraction of oxygen in the mixture]
+phi1 = 0.7;// [fugacity coeffecient of oxygen in the mixture]
+phi2 = 0.85;// [fugacity coeffecient of nitrogen in the mixture]
+phi3 = 0.75;// [fugacity coeffecient of oxygen in the mixture]
+P = 60;// [bar]
+T = 273 + 150;// [K]
+//***********//
+
+phi = exp(x1*log(phi1) + x2*log(phi2) + x3*log(phi3));// [fugacity coeffecient of the mixture]
+f = phi*P;// [bar]
+printf("Fugacity of the gaseous mixture is %.3f bar",f);
\ No newline at end of file diff --git a/647/CH9/EX9.12/Example9_12.sce b/647/CH9/EX9.12/Example9_12.sce new file mode 100755 index 000000000..87ef8d528 --- /dev/null +++ b/647/CH9/EX9.12/Example9_12.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 9.12
+// Page: 356
+
+printf("Example: 9.12 - Page: 356\n\n");
+
+// Solution
+
+//*****Data******//
+T = 372.12;// [K]
+Psat = 100;// [kPa]
+P = 300; //[kPa]
+Vspecific = 1.043*10^(-3);//[cubic m/kg]
+M = 18*10^(-3);// [molecular weight of water, kg/mol]
+R = 8.314;// [J/mol K]
+//***************//
+
+Psat = Psat/100;// [bar]
+P = P/100;// [bar]
+Vl = Vspecific*M;// [cubic m/mol]
+// Vapour is assumed to be like an ideal gas.
+phi = 1;
+fsat = Psat*phi;// [bar]
+fl = fsat*exp(Vl*(P - Psat)*10^5/(R*T));// [bar]
+printf("Fugacity of liquid water is %.4f bar",fl);
\ No newline at end of file diff --git a/647/CH9/EX9.13/Example9_13.sce b/647/CH9/EX9.13/Example9_13.sce new file mode 100755 index 000000000..f66e3f61f --- /dev/null +++ b/647/CH9/EX9.13/Example9_13.sce @@ -0,0 +1,21 @@ +clear;
+clc;
+
+// Example: 9.13
+// Page: 357
+
+printf("Example: 9.13 - Page: 357\n\n");
+
+// Solution
+
+//*****Data******//
+Vl = 90.45*10^(-6);// [molar volume of liquid butadiene, cubic m/mol]
+fsat = 4.12;// [bar]
+P = 10;// [bar]
+Psat = 4.12;// [bar]
+T = 313;// [K]
+R = 8.314;// [J/mol K]
+//************//
+
+fl = fsat*exp(Vl*(P - Psat)*10^5/(R*T));// [bar]
+printf("The fugacity of the liquid water is %.4f bar",fl);
\ No newline at end of file diff --git a/647/CH9/EX9.14/Example9_14.sce b/647/CH9/EX9.14/Example9_14.sce new file mode 100755 index 000000000..1bce40b59 --- /dev/null +++ b/647/CH9/EX9.14/Example9_14.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+// Example: 9.14
+// Page: 357
+
+printf("Example: 9.14 - Page: 357\n\n");
+
+// Solution
+
+//*****Data******//
+b = 0.0391;// [cubic dm/mol]
+P1 = 1000;// [atm]
+T = 1000 + 273;// [K]
+R = 0.0892;// [L bar/K mol]
+deff('[Vreal] = f1(P)','Vreal = R*T/P + b');
+deff('[Videal] = f2(P)','Videal = R*T/P');
+//**************//
+
+// We know that:
+// RTlog(f/P) = integral('Vreal - Videal',P,0,P)
+f = P1*exp((1/(R*T))*integrate('f1(P) - f2(P)','P',0,P1));// [atm]
+phi = f/P1;
+printf("The fugacity of the gas is %d atm \n",f);
+printf("The fugacity coeffecient of the gas is %.3f atm",phi);
\ No newline at end of file diff --git a/647/CH9/EX9.15/Example9_15.sce b/647/CH9/EX9.15/Example9_15.sce new file mode 100755 index 000000000..86a578d0d --- /dev/null +++ b/647/CH9/EX9.15/Example9_15.sce @@ -0,0 +1,23 @@ +clear;
+clc;
+
+// Example: 9.15
+// Page: 359
+
+printf("Example: 9.15 - Page: 359\n\n");
+
+// Solution
+
+//*****Data******//
+Vl = 73*10^(-6);// [cubic m/mol]
+P = 275;// [bar]
+Psat = 4.360;// [bar]
+T = 110 + 273;// [K]
+R = 8.314;// [J/mol K]
+//**************//
+
+// Acetone vapour is assumed to behave like ideal gas.
+phi = 1;
+fsat = Psat;// [bar]
+fl = fsat*exp(Vl*(P - Psat)*10^5/(R*T));// [bar]
+printf("Fugacity of liquid butadiene at 313 K & 10 bar is %.3f bar",fl);
\ No newline at end of file diff --git a/647/CH9/EX9.16/Example9_16.sce b/647/CH9/EX9.16/Example9_16.sce new file mode 100755 index 000000000..028a161c6 --- /dev/null +++ b/647/CH9/EX9.16/Example9_16.sce @@ -0,0 +1,24 @@ +clear;
+clc;
+
+// Example: 9.16
+// Page: 362
+
+printf("Example: 9.16 - Page: 362\n\n");
+
+// Solution
+
+//*****Data******//
+V1 = 2.8;// [Volume of Oxygen, L]
+V2 = 19.6;// [Volume of hydrogen, L]
+R = 1.987;// [cal/K mol]
+//**************//
+
+n1 = V1/22.4;// [moles of Oxygen]
+n2 = V2/22.4;// [moles of Hydrogen]
+n = n1 + n2;// [total number of moles]
+x1 = n1/n;// [mole fraction of Oxygen]
+x2 = n2/n;// [mole fraction of Hydrogen]
+// From Eqn. 9.88:
+deltaS_mix = - (R*(x1*log(x1) + x2*log(x2)));// [cal/K mol]
+printf("The entropy change of mixiong is %.3f cal/K mol",deltaS_mix)
\ No newline at end of file diff --git a/647/CH9/EX9.17/Example9_17.sce b/647/CH9/EX9.17/Example9_17.sce new file mode 100755 index 000000000..d8e26be5b --- /dev/null +++ b/647/CH9/EX9.17/Example9_17.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.17
+// Page: 363
+
+printf("Example: 9.17 - Page: 363\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.17 on page number 363 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.17 on page 363 of the book.");
\ No newline at end of file diff --git a/647/CH9/EX9.18/Example9_18.sce b/647/CH9/EX9.18/Example9_18.sce new file mode 100755 index 000000000..66c6a0b23 --- /dev/null +++ b/647/CH9/EX9.18/Example9_18.sce @@ -0,0 +1,26 @@ +clear;
+clc;
+
+// Example: 9.18
+// Page: 364
+
+printf("Example: 9.18 - Page: 364\n\n");
+
+// Solution
+
+//*****Data******//
+n1 = 0.7;// [moles of helium]
+n2 = 0.3;// [moles of argon]
+R = 8.314;// [J/mol K]
+T = 273 + 25;// [K]
+//******************//
+
+n = n1 + n2;// [total moles]
+x1 = n1/n;// [mole fraction of helium]
+x2 = n2/n;// [mole fraction of argon]
+deltaG_mix = n*R*T*(x1*log(x1) + x2*log(x2));// [J]
+printf("The free energy change of mixing is %.2f J\n",deltaG_mix);
+
+// Since the gases are ideal:
+deltaH_mix = 0;// [J]
+printf("The enthalpy change of mixing is %d J\n",deltaH_mix);
\ No newline at end of file diff --git a/647/CH9/EX9.19/Example9_19.sce b/647/CH9/EX9.19/Example9_19.sce new file mode 100755 index 000000000..dd72f7f3b --- /dev/null +++ b/647/CH9/EX9.19/Example9_19.sce @@ -0,0 +1,36 @@ +clear;
+clc;
+
+// Example: 9.19
+// Page: 364
+
+printf("Example: 9.19 - Page: 364\n\n");
+
+// Solution
+
+//*****Data******//
+V = 20;// [Volume of vessel, L]
+V1 = 12;// [Volume of Hydrogen, L]
+V2 = 10;// [Volume of Nitrogen, L]
+P = 1;// [atm]
+T = 298;// [K]
+P1 = 1;// [atm]
+P2 = 1;// [atm]
+R = 0.082;// [L atm/K mol]
+//************//
+
+n1 = P1*V1/(R*T);// [number of moles of Hydrogen]
+n2 = P2*V2/(R*T);// [number of moles of Nitrogen]
+n = n1 + n2;// [total number of moles]
+Pfinal = n*R*T/V;// [atm]
+p1 = Pfinal*n1;// [partial pressure of Hydrogen, atm]
+p2 = Pfinal*n2;// [partial pressure of Nitrogen, atm]
+deltaG_mix = R*T*(n1*log(p1/P1) + n2*log(p2/P2));// [J]
+printf("Free Energy change of mixing is %.2f J\n",deltaG_mix);
+
+// Since mixing is ideal:
+deltaH_mix = 0;// [J]
+printf("Enthalpy change in mixing is %.2f J\n",deltaH_mix);
+
+deltaS_mix = - (deltaG_mix/T);// [J/K]
+printf("Entropy Change in mixing is %.3f J/K\n",deltaS_mix);
\ No newline at end of file diff --git a/647/CH9/EX9.2/Example9_2.sce b/647/CH9/EX9.2/Example9_2.sce new file mode 100755 index 000000000..7ebae0439 --- /dev/null +++ b/647/CH9/EX9.2/Example9_2.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+
+// Example: 9.2
+// Page: 338
+
+printf("Example: 9.2 - Page: 338\n\n");
+
+// Solution
+
+//*****Data******//
+Vol = 2;// [Volume of the mixture, cubic m/mol]
+y1 = 0.4;// [mole fraction of alcohol, cubic m/mol]
+V1_bar = 38.3*10^(-6);// [partial molar volume of alcohol, cubic m/mol]
+V2_bar = 17.2*10^(-6);// [partial molar volume of water, cubic m/mol]
+V1 = 39.21*10^(-6);// [molar volume of alcohol, cubic m/mol]
+V2 = 18*10^(-6);// [molar volume of water, cubic m/mol]
+//*************//
+
+// From Eqn. 9.9:
+V = y1*V1_bar + (1 - y1)*V2_bar;// [molar volume of the solution]
+n = Vol/V;// [number of moles of solution]
+n1 = y1*n;// [number of moles of alcohol required]
+n2 = (1 - y1)*n;// [number of moles of water required]
+V_alcohol = V1*n1;// [Volume of alcohol required, cubic m]
+V_water = V2*n2;// [Volume of water required, cubic m]
+printf("Volume of alcohol required is %.3f cubic m while volume of water required is %.3f cubic m\n",V_alcohol,V_water);
\ No newline at end of file diff --git a/647/CH9/EX9.20/Example9_20.sce b/647/CH9/EX9.20/Example9_20.sce new file mode 100755 index 000000000..eb527625c --- /dev/null +++ b/647/CH9/EX9.20/Example9_20.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.20
+// Page: 367
+
+printf("Example: 9.20 - Page: 367\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.20 on page number 367 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.20 on page 367 of the book.");
\ No newline at end of file diff --git a/647/CH9/EX9.21/Example9_21.sce b/647/CH9/EX9.21/Example9_21.sce new file mode 100755 index 000000000..61be54d13 --- /dev/null +++ b/647/CH9/EX9.21/Example9_21.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.21
+// Page: 373
+
+printf("Example: 9.21 - Page: 373\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.21 on page number 373 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.21 on page 373 of the book.");
\ No newline at end of file diff --git a/647/CH9/EX9.3/Example9_3.sce b/647/CH9/EX9.3/Example9_3.sce new file mode 100755 index 000000000..9174b03b4 --- /dev/null +++ b/647/CH9/EX9.3/Example9_3.sce @@ -0,0 +1,38 @@ +clear;
+clc;
+
+// Example: 9.3
+// Page: 339
+
+printf("Example: 9.3 - Page: 339\n\n");
+
+// Solution
+
+//*****Data******//
+Vol = 2000;// [cubic cm]
+y1_1 = 0.96;// [mass fraction of ethanol in laboratory alcohol]
+y2_1 = 0.04;// [mass fraction of water in laboratory alcohol]
+y1_2 = 0.56;// [mass fracion of ethanol in vodka]
+y2_2 = 0.44;// [mass fraction of water in vodka]
+Vbar_water1 = 0.816;// [cubic cm/g]
+Vbar_ethanol1 = 1.273;// [cubic cm/g]
+Vbar_water2 = 0.953;// [cubic cm/g]
+Vbar_ethanol2 = 1.243;// [cubic cm/g]
+Density_water = 0.997;// [cubic cm/g]
+//***************//
+
+// Solution (i)
+// From Eqn 9.9
+Va = y1_1*Vbar_ethanol1 + y2_1*Vbar_water1;// [Volume of laboratory alcohol, cubic cm/g]
+mass = Vol/Va;// [g]
+// Let Mw be the mass of water added in laboratory alcohol.
+// Material balance on ethanol:
+Mw = mass*y1_1/y1_2 - mass;// [g]
+Vw = Mw/Density_water;// [Volume of water added, cubic cm]
+printf("Mass of water added is %d g\n",Mw);
+
+// Solution (ii)
+Mv = mass + Mw;// [Mass of vodka, g]
+Vv = y1_2*Vbar_ethanol2 + y2_2*Vbar_water2;// [Volume of ethanol, cubic cm/g]
+V_vodka = Vv*Mv;// [Volume of vodka obtained after conversion, cubic cm]
+printf("The volume of vodka obtained after conversion is %.d cubic cm\n",V_vodka);
\ No newline at end of file diff --git a/647/CH9/EX9.4/Example9_4.sce b/647/CH9/EX9.4/Example9_4.sce new file mode 100755 index 000000000..e6e2f9bda --- /dev/null +++ b/647/CH9/EX9.4/Example9_4.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.4
+// Page: 339
+
+printf("Example: 9.4 - Page: 339\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.4 on page number 339 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.4 on page 339 of the book.");
\ No newline at end of file diff --git a/647/CH9/EX9.5/Example9_5.sce b/647/CH9/EX9.5/Example9_5.sce new file mode 100755 index 000000000..678efa7d0 --- /dev/null +++ b/647/CH9/EX9.5/Example9_5.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.5
+// Page: 340
+
+printf("Example: 9.5 - Page: 340\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.5 on page number 340 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.5 on page 340 of the book.");
\ No newline at end of file diff --git a/647/CH9/EX9.6/Example9_6.sce b/647/CH9/EX9.6/Example9_6.sce new file mode 100755 index 000000000..249e553fb --- /dev/null +++ b/647/CH9/EX9.6/Example9_6.sce @@ -0,0 +1,31 @@ +clear;
+clc;
+
+// Example: 9.6
+// Page: 341
+printf("Example - 9.6 and Page number - 341\n\n");
+
+//Given
+T = 25+273.15;// [K]
+P = 1;// [atm]
+// Component 1 = water
+// Component 2 = methanol
+a = -3.2;// [cubic cm/mol] A constant
+V2 = 40.7;// [cubic cm/mol] Molar volume of pure component 2 (methanol)
+// V1_bar = 18.1 + a*x_2^(2)
+
+// From Gibbs-Duhem equation at constant temperature and pressure we have
+// x_1*dV1_bar + x_2*dV2_bar = 0
+// dV2_bar = -(x_1/x_2)*dV1_bar = -(x_1/x_2)*a*2*x_2*dx_2 = -2*a*x_1*dx_2 = 2*a*x_1*dx_1
+
+// At x_1 = 0: x_2 = 1 and thus V2_bar = V2
+// Integrating the above equation from x_1 = 0 to x_1 in the RHS, and from V2_bar = V2 to V2 in the LHS, we get
+// V2_bar = V2 + a*x_1^(2) - Molar volume of component 2(methanol) in the mixture
+
+printf("The expression for the partial molar volume of methanol(2) is\nV2_bar = V2 + a*x_1^(2) [cubic cm/mol]\n\n");
+
+// At infinite dilution, x_2 approach 0 and thus x_1 approach 1, therefore
+x_1 = 1;// Mole fraction of component 1(water) at infinite dilution
+V2_bar_infinite = V2 + a*(x_1^(2));//[cubic cm/mol]
+
+printf("The partial molar volume of methanol at infinite dilution is %.1f cubic cm/mol",V2_bar_infinite);
\ No newline at end of file diff --git a/647/CH9/EX9.7/Example9_7.sce b/647/CH9/EX9.7/Example9_7.sce new file mode 100755 index 000000000..6a863ed28 --- /dev/null +++ b/647/CH9/EX9.7/Example9_7.sce @@ -0,0 +1,49 @@ +clear;
+clc;
+
+// Example: 9.7
+// Page: 342
+
+printf("Example: 9.7 - Page: 342\n\n");
+
+// Solution
+
+//*****Data******//
+// Data = [X1 V*10^6(cubic m/mol)];
+Data = [0 20;0.2 21.5;0.4 24.0;0.6 27.4;0.8 32.0;1 40];
+//************//
+
+scf(1.1);
+plot(Data(:,1),Data(:,2));
+title("Example 9.7");
+xlabel("Mole fraction");
+ylabel("Molar Volume*10^(6)");
+xgrid();
+
+// Solution (i)
+printf("For X1 = 0.5\n");
+// A tangent is drawn to the curve at X1 = 0.5.
+// The intercept at X2 = 0 or X1 = 1, gives V1_bar.
+V1_bar1 = 33.8*10^(-6);// [cubic m/mol]
+// The intercept at X2 = 1 or X1 = 0, gives V2_bar.
+V2_bar1 = 17*10^(-6);// [cubic m/mol]
+printf("Partial molar volume of component 1 is %.2e cubic m/mol\n",V1_bar1);
+printf("Partial molar volume of component 2 is %.2e cubic m/mol\n",V2_bar1);
+printf("\n");
+
+// Solution (ii)
+printf("For X2 = 0.75\n");
+// A tangent is drawn to the curve at X1 = 0.75.
+// The intercept at X2 = 0 or X1 = 1, gives V1_bar.
+V1_bar2 = 36.6*10^(-6);// [cubic m/mol]
+// The intercept at X2 = 1 or X1 = 0, gives V2_bar.
+V2_bar2 = 12.4*10^(-6);// [cubic m/mol]
+point1 = [0 V1_bar1; 1 V2_bar1];
+point2 = [0 V1_bar2;1 V2_bar2];
+scf(2);
+plot(point1(:,1),point1(:,2),point2(:,1),point2(:,2));
+legend("X1 = 0.5","X1 = 0.75");
+xlabel("Mole fraction");
+ylabel("Molar Volume");
+printf("Partial molar volume of component 1 is %.2e cubic m/mol\n",V1_bar);
+printf("Partial molar volume of component 2 is %.2e cubic m/mol\n",V2_bar);
\ No newline at end of file diff --git a/647/CH9/EX9.8/Example9_8.sce b/647/CH9/EX9.8/Example9_8.sce new file mode 100755 index 000000000..1b94ddfab --- /dev/null +++ b/647/CH9/EX9.8/Example9_8.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.8
+// Page: 352
+
+printf("Example: 9.8 - Page: 352\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.8 on page number 352 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.8 on page 352 of the book.");
\ No newline at end of file diff --git a/647/CH9/EX9.9/Example9_9.sce b/647/CH9/EX9.9/Example9_9.sce new file mode 100755 index 000000000..39d670305 --- /dev/null +++ b/647/CH9/EX9.9/Example9_9.sce @@ -0,0 +1,13 @@ +clear;
+clc;
+
+// Example: 9.9
+// Page: 352
+
+printf("Example: 9.9 - Page: 352\n\n");
+
+// Mathematics is involved in proving but just that no numerical computations are involved.
+// For prove refer to this example 9.9 on page number 352 of the book.
+
+printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n");
+printf(" For prove refer to this example 9.9 on page 352 of the book.");
\ No newline at end of file |